The law of conservation of energy states that the energy of a closed system is constant. Like momentum, energy can't be reduced or increased, but energy does change form.

k = N/m
m = kg

Question: What energies are being traded in the spring and mass system?

answer

elastic potential energy and kinetic energy

Question: When is the velocity the highest?

answer

The kinetic energy and velocity are highest when the spring energy is lowest, at the point of equilibrium. (The dotted line)

Question: How does the total energy change as the spring oscillates.

answer

The total energy is always the same for the same mass and spring constant.

With an actual spring system the energy would eventually change forms into thermal energy, but this simulation doesn't include heat loss.

The total energy is always the same value. This means we can build an equation that sets the total energy at one time equal to the total energy at any other time.

Energy cannot be created or destroyed, but it can change forms.

$$E_i = E_f$$

$ E_i $ = initial energy [J]
$ E_f $ = final energy [J]

$$K + U_g + U_s = K + U_g + U_s$$

$ K $ = kinetic energy [J]
$ U_g $ = gravitational potential energy [J]
$ U_s $ = elastic potential energy [J]

We've only covered 3 types of energy. As we learn more types the equation can gain new terms. We don't have to include every term in the equation. Only include types of energy that change.

If an object changes speed, include kinetic energy.

If an object moves vertically, include gravitational potential energy.

If a spring is compressed, include spring potential energy.

For now, we will assume no thermal energy loss, but we will explore it in the next section, thermodynamics.

Example: Build a conservation of energy equation for dropping a ball.

strategy

This equation only needs to include kinetic and gravitational potential energy. Initial kinetic energy is zero because the ball starts at rest. Choosing the final height to be zero will set the final gravitational potential energy to zero.

solution

$$U_g = K$$ $$mgh = \frac{1}{2}mv^2$$ $$\cancel{m}gh = \frac{1}{2} \cancel{m}v^2$$ $$gh = \frac{1}{2}v^2$$ $$\boxed{v^2 = 0+2gh}$$

This energy equation happens to be the same as one of the equations of motion for constant acceleration.

$$v^2 = u^2 + 2a\Delta x$$

Example: A ball falls from rest off a 0.80 m table. Before we solve this problem, what final and initial energies can we set to zero?

answer

The velocity is at rest initially. This means that the initial kinetic energy will be zero as well.

We can choose to make height zero at the bottom of the table. This means there is no final gravitational potential energy.

How fast is the ball going just before it hits the ground?

solution

$$K_i + U_i = K_f + U_f$$ $$0+U_g = K+0$$ $$mgh = \tfrac{1}{2}mv^{2}$$ $$m(9.8)(0.80)=\tfrac{1}{2}mv^2$$ $$7.84m=0.5mv^2$$ $$7.84=0.5v^2$$ $$15.68=v^2$$ $$\pm 3.959\, \mathrm{ \tfrac{m}{s} } = v$$

Example: If you throw a ball straight up at 10 m/s, how high will it go?

solution

$$K = U_g$$ $$\tfrac{1}{2}mv^{2} = mgh$$ $$\tfrac{1}{2}m(10)^2=m(9.8)h$$ $$50m=9.8mh$$ $$50=9.8h$$ $$5.10 \, \mathrm{m} = h$$

Question: Which frictionless slope will give you the highest speed at the bottom? Explain your choice.

answer

Each path starts with the same gravitational potential energy. This energy is converted to the same kinetic energy.

They will all have the same speed, but in different directions.

Example: A 0.43 kg soccer ball kicked at 10 m/s rolls down a 30 m tall hill. How fast will the ball be moving at the bottom of the hill?

solution

Some of the energy is converted into heat through friction, and into rotational kinetic energy. We can't track those energy types so it will limit the accuracy of our answer.

$$U_{g} + K_{i} = K_{f}$$ $$mgh + \tfrac{1}{2}mu^{2} = \tfrac{1}{2}mv^{2}$$ $$(0.43)(9.8)(30)+\tfrac{1}{2}(0.43)(10)^2=\tfrac{1}{2}(0.43)v^2$$ $$126.42+21.5=0.215v^2$$ $$147.92=0.215v^2$$ $$688=v^2$$ $$\pm 26.2 \mathrm{\tfrac{m}{s}} = v$$

Example: A roller coaster cart starts at rest 328 ft high on the top of the first drop on "Superman: Escape from Krypton" at Six Flags Magic Mountain. How fast is the cart going at the bottom of the hill?

solution

$$328\,\mathrm{ft} \left(\frac{0.3048 \,\mathrm{m}}{1 \,\mathrm{ft}} \right) = 100 \,\mathrm{m}$$
$$U_{g} = K$$ $$mgh = \tfrac{1}{2}mv^{2}$$ $$m(9.8)(100)=\tfrac{1}{2}mv^2$$ $$980m=0.5mv^2$$ $$980=0.5v^2$$ $$980 = \tfrac{1}{2}v^{2}$$ $$\pm 44.3 \, \mathrm{\tfrac{m}{s}} = v$$

Example: A frictionless roller coaster starts from rest at point A. What is the velocity of the roller coaster at points B, C, and D. Assume each grid square is 10 m × 10 m.

solution

Conservation of energy for gravitational and kinetic energy is independent of mass.

$$\text{point A}$$ $$E_A = mgh$$ $$E_A=m(9.8)(100)$$ $$E_A=980m$$ $$\frac{E_A}{m} = 980 \, \mathrm{\tfrac{J}{kg}}$$

Now we can set the energy / mass at point A equal to the energy / mass at points B,C,D.

$$\text{point B}$$ $$E_A = E_B$$ $$980 = gh + \tfrac{1}{2}v^2$$ $$980 = (9.8)(8) + \tfrac{1}{2}v^2$$ $$980 - 78.4 = \tfrac{1}{2}v^2$$ $$1803.2 = v^2$$ $$v = 42.5 \, \mathrm{\tfrac{m}{s}}$$

$$\text{point C}$$ $$980 = (9.8)(47) + \tfrac{1}{2}v^2$$ $$v = 32.2 \, \mathrm{\tfrac{m}{s}}$$

$$\text{point D}$$ $$980 = (9.8)(20) + \tfrac{1}{2}v^2$$ $$v = 39.6 \, \mathrm{\tfrac{m}{s}}$$

Example: A ball on a 1.0 meter long string is let go at a 30° angle. Calculate the velocity at 45° and 90° as it swings back and forth.

convert angles into heights

$$h = (1.0)sin(30) = 0.5 \,\mathrm{m}$$ $$h = (1.0)sin(90) = 1\,\mathrm{m}$$ $$h = (1.0)sin(45) = 0.71\,\mathrm{m}$$

solution 90°

$$K_i + U_i = K_f + U_f$$ $$U_{gi} = K +U_{gf}$$ $$mgh_i = \tfrac{1}{2}mv^2 + mgh_f$$ $$m(9.8)(-0.5)=\tfrac{1}{2}mv^2+m(9.8)(-1)$$ $$-4.9m=0.5mv^2-9.8m$$ $$-4.9=0.5v^2-9.8$$ $$(9.8)(0.5) = \tfrac{1}{2}v^2$$ $$3.13 \, \tfrac{m}{s} = v $$

solution 45°

$$K_i + U_i = K_f + U_f$$ $$U_{gi} = K +U_{gf}$$ $$mgh_i = \tfrac{1}{2}mv^2 + mgh_f$$ $$m(9.8)(-0.5)=\tfrac{1}{2}mv^2+m(9.8)(-0.71)$$ $$-4.9m=0.5mv^2-6.958m$$ $$-4.9=0.5v^2-6.958$$ $$(9.8)(0.21) = \tfrac{1}{2}v^2$$ $$2.03 \, \tfrac{m}{s} = v $$

Example: A 0.20 kg ball is placed on a spring compressed down 0.15 m. The internet says that your spring has a spring constant of 200.0 N/m.

How fast should the ball be moving right after it leaves the spring?

solution

If we define h to be zero at the spring equilibrium, then h=x.

$$U_s + U_g = K$$ $$\tfrac{1}{2}kx^{2} + mgh= \tfrac{1}{2}mv^{2}$$ $$\tfrac{1}{2}kx^{2} + mg x= \tfrac{1}{2}mv^{2}$$ $$\tfrac{1}{2}(200)(-0.15)^{2} + (0.2)(9.8)(-0.15)= \tfrac{1}{2}(0.2)v^{2}$$ $$1.956 = 0.1v^{2}$$ $$4.42 \, \mathrm{\tfrac{m}{s}} = v$$

How high will the ball go?

solution

$$K = U_g$$ $$\tfrac{1}{2}mv^{2} = mgh$$

We know from the previous question that the total energy is 1.956 J.

$$1.956 = (0.20)(9.8)h$$ $$1.956 = 1.96h$$ $$1.00 \, \mathrm{m} = h$$

dampening: none full

k = kg/s²
m = kg

Example: How fast is the mass moving at the spring equilibrium (x = 0)?
Use the default values.

solution

We will define h = 0 at x = 0. This makes x = h.

$$\tfrac{1}{2}mv^2 + mgh + \tfrac{1}{2}kx^2$$ $$\tfrac{1}{2}mv^2 + mgx + \tfrac{1}{2}kx^2$$

The initial energy will be when the spring and height are at equilibrium (x = 0). The final energy will be at the lowest point, when the spring is at rest (v = 0).

$$v_i = \, ? \quad x_i =0 \quad \quad \quad \quad v_f = 0 \quad x_f = -150$$ $$\tfrac{1}{2}mv^2 + mgx + \tfrac{1}{2}kx^2 = \tfrac{1}{2}mv^2 + mgx + \tfrac{1}{2}kx^2$$ $$\tfrac{1}{2}mv^2 = mgx + \tfrac{1}{2}kx^2$$ $$\tfrac{1}{2}(13)v^2 = (13)(9.8)(-150) + \tfrac{1}{2}(10)(-150)^2$$ $$6.5v^2 = -19110 + 112500$$ $$6.5v^2 = 93390$$ $$v = \pm 119.9 \tfrac{m}{s}$$

Example: When the dampening is turned on, the spring system slowly settles on a single position. Write an equation to predict that position in terms of the mass and spring constant.

strategy

To build the equation, solve Newton's second law for an acceleration of zero.

solution

$$\sum F = ma $$ $$F_s-F_g = ma $$ $$-kx-mg = 0 $$ $$-kx = mg $$ $$\boxed{x = -\frac{mg}{k} }$$

Is this equation correct? Test it out with the simulation.

Click to Run

Open the PhET simulation, switch to the energy lab mode on the bottom right.

Investigation: What is the value of the spring constant?
Use Newton's second law to find the solution.

solution

We need to solve a Newton's second law equation. Add a known mass to the hook and click stop to get the velocity and acceleration to zero.

$$\sum F = ma$$ $$F_s-F_g = ma$$ $$kx-mg = 0$$ $$kx=mg$$

We can read mass and gravity directly from the simulation. We can measure the displacement with the ruler. This is easier if you check the displacement box in the top right.

$$m=0.100 \,\mathrm{kg} \quad g=9.8\,\mathrm{\tfrac{m}{s^2}} \quad x=0.17\,\mathrm{m} $$ $$k(0.17)=(0.100)(9.8)$$ $$0.17k=0.98$$ $$k=\frac{0.98}{0.17}$$ $$k=5.8 \, \mathrm{\tfrac{kg}{s^2}}$$

Investigation: What is the value of the spring constant?
Use conservation of energy to find the solution.

solution

Conservation of energy requires two different moments in time. This simulation makes velocity impossible to measure, so we must choose two moments where v = 0.

Point one will be at spring equilibrium. Place the 100 g mass on the hook and let it get exactly at the point of spring equilibrium. This is easier if you check the displacement box in the top right.

Point two will be at the bottom of the springs drop. Make sure dampening is off so we don't lose energy to heat.

$$K_i + U_{gi} + U_{si} = K_f + U_{gf} + U_{sf}$$

The velocity is zero at both moments. If we define height to be zero at the spring equilibrium, then both potential energies are zero at the start.

$$0 = U_g + U_s$$ $$0 = mgx + \tfrac{1}{2}kx^2$$ $$0 = mgx + \tfrac{1}{2}kx^2$$

We can read mass and gravity directly from the simulation. We can measure the displacement with the ruler. This is easier if you check the displacement box in the top right. Also slow motion and pause help.

$$m=0.100 \,\mathrm{kg} \quad g=9.8\,\mathrm{\tfrac{m}{s^2}} \quad x=-0.34\,\mathrm{m} $$ $$0=(0.100)(9.8)(-0.34)+\tfrac{1}{2}k(-0.34)^2$$ $$0=-0.3332+0.0578k$$ $$0.3332=0.0578k$$ $$k=\frac{0.3332}{0.0578}$$ $$k=5.8 \, \mathrm{\tfrac{kg}{s^2}}$$

Investigation: Calculate the mass of the red and blue weights.

solution

We can just reuse our method for getting the spring constant, but for mass.

$$kx=mg$$

Place each mass on the hook with dampening on high, and measure how much it displaces the spring.

$$\text{blue mass}$$ $$(5.8)(0.38)=m(9.8)$$ $$2.204=9.8m$$ $$m=\frac{2.204}{9.8}$$ $$m=0.224\, \mathrm{kg}$$

$$\text{red mass}$$ $$(5.8)(0.61)=m(9.8)$$ $$3.538=9.8m$$ $$m=\frac{3.538}{9.8}$$ $$m=0.361\, \mathrm{kg}$$

Perfectly Elastic Collisions

When objects collide the total momentum is conserved. Normally some of the kinetic energy of the objects will be converted into thermal or rotational energy, but in some situations the energy stays kinetic. A collision that doesn't lose any kinetic energy is called perfectly elastic. This means that for perfectly bouncy collisions we can use two conservation equations.

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ $$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$

With two equations we can solve for two unknowns. We get these equations if we substitute one equation into the other and solve for the final velocities.

$$v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 + \frac{2m_2}{m_1+m_2}u_2$$ $$v_2 = \frac{2m_1}{m_1+m_2}u_1 + \frac{m_2-m_1}{m_1+m_2}u_2$$

These equations are only approximations at our human scale, but at the atomic scale perfectly elastic collisions aren't unusual. So you'd get very accurate results for gas or liquid atoms colliding.

Example: A 2 kg ball moving at 2 m/s to the right collides with a 7 kg ball at rest. Approximate the final velocities of each ball assuming that no kinetic energy is lost.

solution

$$2\, \mathrm{kg \, ball}$$ $$v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 + \frac{2m_2}{m_1+m_2}u_2$$ $$v_1 = \left(\frac{2-7}{2+7} \right)2 + \left(\frac{2(7)}{2+7} \right)0$$ $$v_1 = \left(\frac{2-7}{2+7} \right)2$$ $$v_1 = -1.\overline{1} \, \mathrm{\tfrac{m}{s}} $$
$$7\, \mathrm{kg \, ball}$$ $$v_2 = \frac{2m_1}{m_1+m_2}u_1 + \frac{m_2-m_1}{m_1+m_2}u_2$$ $$v_2 = \left(\frac{2(2)}{2+7}\right)2 + \left(\frac{7-2}{2+7}\right)0$$ $$v_2 = \left(\frac{2(2)}{2+7}\right)2$$ $$v_2 = 0.\overline{8} \, \mathrm{\tfrac{m}{s}} $$

If the 2 colliding objects have the same mass we can simplify the equations.

$$v_1=u_2 \quad \quad v_2=u_1 $$

After an elastic collision objects with the same mass trade velocities.

Example: A billiards ball is typically 0.16 kg. Although the cue ball is normally a bit heavier. The 8 ball moving at 12 m/s collides with the 3 ball at rest. Approximate the final velocities of each ball assuming no kinetic energy loss.

solution

Not much math here. They just trade velocities.
The 8 ball is now at rest and the 3 ball is moving at 12 m/s.

practice problems (25)

In case you wanted more practice I used AI to make some more problems. The rest of the site I made by hand, but generating endless problems seemed safe. I did find mistakes in the AI generated problems, and there are probably some I didn't find. Let me know if something could be fixed. I also added a practice problem on each page with no solution. That's intentional. Have fun!

Assume Earth gravity and no air resistance unless the problem says otherwise.

printout.pdf show topics and mark harder problems

Example: A 2.0 kg cart rolls down a curved, frictionless track and drops 1.5 m in height. How much gravitational potential energy does it lose, and how much kinetic energy does it gain?

solution

Choose the starting height as zero, so the cart ends at -1.5 m.

$$\Delta U_g = mgh_f - mgh_i$$ $$\Delta U_g = (2.0)(9.8)(-1.5) - 0$$ $$\Delta U_g = -29.4\,\mathrm{J}$$

Energy is conserved, so the 29.4 J of gravitational potential energy the cart loses becomes 29.4 J of kinetic energy. The curve of the track doesn't matter, only the change in height.

Example: A spring launcher pushes a cart along a level track. The spring's energy decreases by 18 J. How much kinetic energy does the cart gain?

solution

The cart stays at the same height, so only spring energy and kinetic energy change.

$$K_i + U_{si} = K_f + U_{sf}$$ $$K_f - K_i = U_{si} - U_{sf}$$ $$\Delta K = 18\,\mathrm{J}$$

Every joule the spring loses becomes kinetic energy of the cart.

Example: A skateboarder is moving at 6.0 m/s at the bottom of a smooth hill. Ignoring friction, how high up the hill can the skateboarder coast before stopping?

solution

Choose the bottom of the hill as zero height. At the highest point the speed is zero.

$$K_i + U_{gi} = K_f + U_{gf}$$ $$\tfrac{1}{2}mu^{2} + 0 = 0 + mgh$$ $$\tfrac{1}{2}\cancel{m}(6.0)^{2} = \cancel{m}(9.8)h$$ $$18 = 9.8h$$ $$h = \frac{18}{9.8}$$ $$h = 1.84\,\mathrm{m}$$

The mass cancels, so a heavier skateboarder would coast to the same height.

Example: A 4.0 kg cart moving at 5.0 m/s rolls up a frictionless track. How much kinetic energy does it still have at a point 0.80 m higher?

solution

$$K_i + U_{gi} = K_f + U_{gf}$$ $$\tfrac{1}{2}(4.0)(5.0)^{2} + 0 = K_f + (4.0)(9.8)(0.80)$$ $$50 = K_f + 31.4$$ $$K_f = 18.6\,\mathrm{J}$$

The cart traded 31.4 J of kinetic energy for gravitational potential energy, and it is still moving.

Example: A 1500 g toy car starts from rest at the top of a track that is 0.90 m tall. Ignoring friction, how fast is it moving at the bottom?

solution

The mass cancels, so we don't need to convert grams to kilograms.

$$K_i + U_{gi} = K_f + U_{gf}$$ $$0 + mgh = \tfrac{1}{2}mv^{2} + 0$$ $$\cancel{m}(9.8)(0.90) = \tfrac{1}{2}\cancel{m}v^{2}$$ $$8.82 = 0.5v^{2}$$ $$v^{2} = 17.6$$ $$v = 4.20\,\mathrm{\tfrac{m}{s}}$$

Example: A cart is moving at 8.0 m/s at the bottom of a frictionless track. What is its speed after it climbs 2.0 m higher?

solution

$$K_i + U_{gi} = K_f + U_{gf}$$ $$\tfrac{1}{2}mu^{2} + 0 = \tfrac{1}{2}mv^{2} + mgh$$ $$\tfrac{1}{2}\cancel{m}(8.0)^{2} = \tfrac{1}{2}\cancel{m}v^{2} + \cancel{m}(9.8)(2.0)$$ $$32 = 0.5v^{2} + 19.6$$ $$12.4 = 0.5v^{2}$$ $$v^{2} = 24.8$$ $$v = 4.98\,\mathrm{\tfrac{m}{s}}$$

We didn't need the mass, because every term has m in it.

Example: A 2.0 kg cart starts from rest and rolls down a frictionless ramp that is 5.0 m long. How fast is it moving at the bottom?

solution

This cannot be solved from the information given. Conservation of energy needs the change in height, not the length of the ramp. A 5.0 m ramp could be almost flat or very steep. The angle of the ramp or its height is missing.

Example: A 0.30 kg cart is launched by a spring compressed 15 cm. The spring constant is 120 N/m. What speed does the cart have when it leaves the spring on a level track?

solution

$$15\,\mathrm{cm} = 15(0.01)\,\mathrm{m}$$ $$15\,\mathrm{cm} = 0.15\,\mathrm{m}$$
$$U_{si} = K_f$$ $$\tfrac{1}{2}kx^{2} = \tfrac{1}{2}mv^{2}$$ $$\tfrac{1}{2}(120)(0.15)^{2} = \tfrac{1}{2}(0.30)v^{2}$$ $$1.35 = 0.15v^{2}$$ $$v^{2} = 9.0$$ $$v = 3.0\,\mathrm{\tfrac{m}{s}}$$

Example: A 0.60 kg cart moving at 2.0 m/s runs into a 150 N/m spring bumper on a level frictionless track. How far does the spring compress before the cart stops?

solution

$$K_i = U_{sf}$$ $$\tfrac{1}{2}mu^{2} = \tfrac{1}{2}kx^{2}$$ $$\tfrac{1}{2}(0.60)(2.0)^{2} = \tfrac{1}{2}(150)x^{2}$$ $$1.2 = 75x^{2}$$ $$x^{2} = 0.016$$ $$x = 0.126\,\mathrm{m}$$

That's about 13 cm. After the cart stops, the spring pushes it back out at 2.0 m/s.

Example: A spring with k = 80 N/m is compressed 0.25 m. It launches a 0.50 kg cart up a frictionless ramp. How high does the cart rise before it stops?

solution

$$U_{si} = U_{gf}$$ $$\tfrac{1}{2}kx^{2} = mgh$$ $$\tfrac{1}{2}(80)(0.25)^{2} = (0.50)(9.8)h$$ $$2.5 = 4.9h$$ $$h = 0.510\,\mathrm{m}$$

Example: A pole vaulter sprints down the runway at 9.0 m/s. If all of that kinetic energy became gravitational potential energy, how high could the vaulter raise their center of mass?

solution

$$K_i = U_{gf}$$ $$\tfrac{1}{2}\cancel{m}v^{2} = \cancel{m}gh$$ $$\tfrac{1}{2}(9.0)^{2} = (9.8)h$$ $$40.5 = 9.8h$$ $$h = 4.13\,\mathrm{m}$$

Elite vaulters clear about 6 m. Their center of mass already starts about 1 m off the ground, and they push up on the pole with their arms at the top, adding a little more energy.

Example: A 500 kg roller coaster car starts from rest at point A, 20 m above the ground. Ignoring friction, fill in the table for points B and C.

point	height (m)	U_g (J)	K (J)	speed (m/s)
A	20	?	0	0
B	5	?	?	?
C	12	?	?	?

solution

$$E_A = mgh$$ $$E_A = (500)(9.8)(20)$$ $$E_A = 98\,000\,\mathrm{J}$$

The total energy stays 98 000 J at every point. Subtract the gravitational potential energy to find the kinetic energy.

$$\text{point B}$$ $$U_g = (500)(9.8)(5)$$ $$U_g = 24\,500\,\mathrm{J}$$ $$K = 98\,000 - 24\,500$$ $$K = 73\,500\,\mathrm{J}$$ $$73\,500 = \tfrac{1}{2}(500)v^{2}$$ $$v = 17.1\,\mathrm{\tfrac{m}{s}}$$

$$\text{point C}$$ $$U_g = (500)(9.8)(12)$$ $$U_g = 58\,800\,\mathrm{J}$$ $$K = 98\,000 - 58\,800$$ $$K = 39\,200\,\mathrm{J}$$ $$39\,200 = \tfrac{1}{2}(500)v^{2}$$ $$v = 12.5\,\mathrm{\tfrac{m}{s}}$$

The lower the car goes, the faster it moves.

Example: A 70 kg skier starts from rest at the top of a 12 m tall hill. Ignoring friction, how fast is the skier moving at the bottom? Why doesn't the mass matter?

solution

$$U_{gi} = K_f$$ $$mgh = \tfrac{1}{2}mv^{2}$$ $$(70)(9.8)(12) = \tfrac{1}{2}(70)v^{2}$$ $$8232 = 35v^{2}$$ $$v^{2} = 235$$ $$v = 15.3\,\mathrm{\tfrac{m}{s}}$$

A heavier skier starts with more gravitational potential energy, but also needs more kinetic energy to reach the same speed. The two effects cancel exactly.

Example: A 0.40 kg ball is moving at 3.0 m/s when it is 1.2 m above the floor. Ignoring air resistance, how fast is it moving when it reaches the floor?

solution

The ball starts with both kinetic and gravitational potential energy. Choose the floor as zero height.

$$K_i + U_{gi} = K_f + U_{gf}$$ $$\tfrac{1}{2}mu^{2} + mgh = \tfrac{1}{2}mv^{2} + 0$$ $$\tfrac{1}{2}(0.40)(3.0)^{2} + (0.40)(9.8)(1.2) = \tfrac{1}{2}(0.40)v^{2}$$ $$1.8 + 4.70 = 0.20v^{2}$$ $$6.50 = 0.20v^{2}$$ $$v^{2} = 32.5$$ $$v = 5.70\,\mathrm{\tfrac{m}{s}}$$

The direction the ball was moving at 3.0 m/s doesn't matter. Energy is a scalar.

Question: From the top of a cliff, you throw one rock straight up at 10 m/s and an identical rock straight down at 10 m/s. Ignoring air resistance, which rock is moving faster when it hits the ground below?

answer

They hit the ground at the same speed.

Both rocks start at the same height with the same speed, so they start with the same total energy. Energy is a scalar, so the direction of the throw doesn't matter. The rock thrown up rises, stops, and falls back past the cliff edge moving down at 10 m/s, just like the other rock.

Example: A cart moves at 7.0 m/s at the bottom of a frictionless track. It reaches the top of a hill still moving at 3.0 m/s. How tall is the hill?

solution

$$K_i + U_{gi} = K_f + U_{gf}$$ $$\tfrac{1}{2}mu^{2} + 0 = \tfrac{1}{2}mv^{2} + mgh$$ $$\tfrac{1}{2}\cancel{m}(7.0)^{2} = \tfrac{1}{2}\cancel{m}(3.0)^{2} + \cancel{m}(9.8)h$$ $$24.5 = 4.5 + 9.8h$$ $$20 = 9.8h$$ $$h = 2.04\,\mathrm{m}$$

Example: A 0.50 kg glider moving at 2.0 m/s runs into a spring bumper and stops after compressing it 0.20 m. What is the spring constant?

solution

$$K_i = U_{sf}$$ $$\tfrac{1}{2}mu^{2} = \tfrac{1}{2}kx^{2}$$ $$\tfrac{1}{2}(0.50)(2.0)^{2} = \tfrac{1}{2}k(0.20)^{2}$$ $$1.0 = 0.020k$$ $$k = 50\,\mathrm{\tfrac{N}{m}}$$

Example: A 1.2 kg ball is thrown straight up at 5.0 m/s. A motion sensor shows that the ball stops rising after 0.51 s. Use kinematics to find how high it rises, then check your answer with conservation of energy.

solution

$$\text{kinematics}$$ $$\Delta y = u\Delta t + \tfrac{1}{2}a\Delta t^{2}$$ $$\Delta y = (5.0)(0.51) + \tfrac{1}{2}(-9.8)(0.51)^{2}$$ $$\Delta y = 1.28\,\mathrm{m}$$

$$\text{energy}$$ $$K_i = U_{gf}$$ $$\tfrac{1}{2}mu^{2} = mgh$$ $$\tfrac{1}{2}(1.2)(5.0)^{2} = (1.2)(9.8)h$$ $$15 = 11.76h$$ $$h = 1.28\,\mathrm{m}$$

The two methods agree. The energy method didn't need the time.

Example: A 900 kg car is moving at 18 km/h. If all of its kinetic energy could be stored in a giant spring bumper with k = 20 000 N/m, how far would the spring compress?

solution

$$18\,\mathrm{\tfrac{\textcolor{DeepPink}{km}}{\textcolor{DodgerBlue}{h}}}\left(\frac{1000\,\mathrm{m}}{1\,\textcolor{DeepPink}{\mathrm{km}}}\right)\left(\frac{1\,\textcolor{DodgerBlue}{\mathrm{h}}}{3600\,\mathrm{s}}\right)$$ $$5.0\,\mathrm{\tfrac{m}{s}}$$
$$K_i = U_{sf}$$ $$\tfrac{1}{2}mu^{2} = \tfrac{1}{2}kx^{2}$$ $$\tfrac{1}{2}(900)(5.0)^{2} = \tfrac{1}{2}(20\,000)x^{2}$$ $$11\,250 = 10\,000x^{2}$$ $$x^{2} = 1.125$$ $$x = 1.06\,\mathrm{m}$$

Even at a slow 18 km/h, a car needs a very stiff spring and more than a meter to stop gently.

Example: A 0.20 kg cart starts at rest against a spring compressed 10 cm. The spring constant is 180 N/m. The cart leaves the spring and climbs a frictionless track. What is its speed after it rises 0.50 m?

solution

$$10\,\mathrm{cm} = 10(0.01)\,\mathrm{m}$$ $$10\,\mathrm{cm} = 0.10\,\mathrm{m}$$
$$U_{si} = K_f + U_{gf}$$ $$\tfrac{1}{2}kx^{2} = \tfrac{1}{2}mv^{2} + mgh$$ $$\tfrac{1}{2}(180)(0.10)^{2} = \tfrac{1}{2}(0.20)v^{2} + (0.20)(9.8)(0.50)$$ $$0.90 = 0.10v^{2} + 0.98$$ $$-0.08 = 0.10v^{2}$$

A squared speed can't be negative, so the cart never reaches 0.50 m. The spring only stores 0.90 J, but rising 0.50 m takes 0.98 J. The cart's highest point is:

$$0.90 = (0.20)(9.8)h$$ $$h = 0.46\,\mathrm{m}$$

Example: A 0.50 kg cart starts at rest against a spring compressed 20 cm. The spring constant is 250 N/m. The cart leaves the spring, climbs a frictionless hill, and is still moving at 3.0 m/s at the top. How tall is the hill?

solution

$$20\,\mathrm{cm} = 20(0.01)\,\mathrm{m}$$ $$20\,\mathrm{cm} = 0.20\,\mathrm{m}$$
$$U_{si} = K_f + U_{gf}$$ $$\tfrac{1}{2}kx^{2} = \tfrac{1}{2}mv^{2} + mgh$$ $$\tfrac{1}{2}(250)(0.20)^{2} = \tfrac{1}{2}(0.50)(3.0)^{2} + (0.50)(9.8)h$$ $$5.0 = 2.25 + 4.9h$$ $$2.75 = 4.9h$$ $$h = 0.561\,\mathrm{m}$$

Example: A 0.30 kg block slides from rest down a frictionless track, dropping 1.4 m in height. At the bottom it runs into a 400 N/m spring. How far does the spring compress when the block first stops?

solution

At the end the block is stopped, so all the lost gravitational potential energy is stored in the spring.

$$U_{gi} = U_{sf}$$ $$mgh = \tfrac{1}{2}kx^{2}$$ $$(0.30)(9.8)(1.4) = \tfrac{1}{2}(400)x^{2}$$ $$4.12 = 200x^{2}$$ $$x^{2} = 0.0206$$ $$x = 0.143\,\mathrm{m}$$

Example: A 0.010 kg bullet is fired into a 2.0 kg wooden block hanging from strings. The bullet stays in the block, and the block swings up 0.32 m. How fast was the bullet moving?

solution

Work backward. Use conservation of energy for the swing, then conservation of momentum for the collision.

$$\text{swing}$$ $$K_i = U_{gf}$$ $$\tfrac{1}{2}\cancel{m}v^{2} = \cancel{m}gh$$ $$\tfrac{1}{2}v^{2} = (9.8)(0.32)$$ $$v^{2} = 6.27$$ $$v = 2.50\,\mathrm{\tfrac{m}{s}}$$
$$\text{collision}$$ $$m_1u_1 + m_2u_2 = (m_1 + m_2)v$$ $$(0.010)u_1 + (2.0)(0) = (2.010)(2.50)$$ $$0.010u_1 = 5.03$$ $$u_1 = 503\,\mathrm{\tfrac{m}{s}}$$

Kinetic energy isn't conserved in the collision, because the bullet sticking in the wood produces heat. That's why we use momentum for the collision and energy only for the swing.

Example: Two equal-mass hockey pucks collide elastically on frictionless ice. Puck A moves right at 4.0 m/s, and puck B moves left at 1.0 m/s. What are their velocities after the collision?

solution

In an elastic collision between equal masses, the objects trade velocities.

$$v_A = -1.0\,\mathrm{\tfrac{m}{s}}$$ $$v_B = 4.0\,\mathrm{\tfrac{m}{s}}$$

Puck A now moves left at 1.0 m/s, and puck B moves right at 4.0 m/s.

Example: A 2.0 kg cart moving right at 4.0 m/s collides elastically with a 6.0 kg cart at rest. After the collision, the 2.0 kg cart bounces left at 2.0 m/s. Use conservation of momentum to find the 6.0 kg cart's velocity, then check that kinetic energy is conserved.

solution

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ $$(2.0)(4.0) + (6.0)(0) = (2.0)(-2.0) + (6.0)v_2$$ $$8.0 = -4.0 + 6.0v_2$$ $$v_2 = 2.0\,\mathrm{\tfrac{m}{s}}$$

$$\text{before}$$ $$K = \tfrac{1}{2}(2.0)(4.0)^{2}$$ $$K = 16\,\mathrm{J}$$

$$\text{after}$$ $$K = \tfrac{1}{2}(2.0)(-2.0)^{2} + \tfrac{1}{2}(6.0)(2.0)^{2}$$ $$K = 4.0 + 12$$ $$K = 16\,\mathrm{J}$$

The kinetic energy is the same before and after, so the collision really is elastic. You can also check with the elastic collision equations on this page.

Reading (6 minutes): Read Explainer: Kinetic and potential energy from Science News Explores. Then answer these questions.

Why does doubling an object's speed make its kinetic energy four times larger rather than two times larger?

answer

Kinetic energy depends on speed squared. Doubling the speed multiplies the squared speed by four.

A skateboarder rolls up a ramp and slows down. What energy change is taking place, and what happens as the skateboarder rolls back down?

answer

As the skateboarder rises, kinetic energy changes into gravitational potential energy. On the way down, gravitational potential energy changes back into kinetic energy.

The article compares doubling mass with doubling speed. Which change has the larger effect on kinetic energy, and why?

answer

Doubling speed has the larger effect because it makes kinetic energy four times larger. Doubling mass only doubles kinetic energy.