Work and Kinetic Energy

Part A = It is equal to the kinetic energy of the lighter block.

Part B = twice as fast

Part C = The heavy block must be pushed 4 times farther than the light block.

Solutions Below:

Part A

A. It is smaller than the kinetic energy of the lighter block.

B. It is equal to the kinetic energy of the lighter block.

C. It is larger than the kinetic energy of the lighter block.

D. It cannot be determined without knowing the force and the mass of each block.

Let’s do a comparison with some example values and then I’ll explain the theory. Let’s say that the first block is 10kg and the second block is 40kg (4 times heavier). Say that the force is 100N. Since F = ma, we know that:

For the first block:

F = ma

100N = 10kg * a

a_{1} = 10m/s^{2}

For the second block:

F = ma

100N = 40kg * a

a_{2} = 2.5m/s^{2}

Now say that the distance is 10m. We can find the final velocity of each block via:

V_{f}^{2} = V_{0}^{2} + 2a(d)

This is just one of the kinematic equations that you should know from previous chapters. So the final velocities are (assume each block starts from zero):

Block 1 = 14.14m/s

Block 2 = 7.07m/s

There is a pattern here, I’ll explain at the end if you don’t see it already.

Now that we have the final velocities, we can solve for the kinetic energy of each block, using the equation KE = 1/2mv^{2}:

Block 1:

KE = 1/2mv^{2}

KE = 1/2(10)(14.14)^{2}

KE = 1,000J

Block 2:

KE = 1/2mv^{2}

KE = 1/2(40)(7.07)^{2}

KE = 1,000J

So the answer is B. It is equal to the kinetic energy of the lighter block.

Let me elaborate a little bit. There is something called the work-energy theorem that basically says that the change in kinetic energy equals the net work done on an object. Since work = force times distance we know that both objects (since the same force is applied over the same distance) will end up with the same kinetic energy. But you may have also noticed a pattern. That is, that when the mass of one block is 4 times the other, acceleration is only twice as much (or for the other block, half). And kinetic energy is half of that (one to one). The reason of course is that because acceleration and kinetic energy involve square units. Acceleration is m/s^{2} and KE squares velocity. Anyway, if that doesn’t make sense, don’t worry about it, you can solve this type of problem by using the formulas posted here and remembering the work-energy theorem.

To recap, the answer is:

B. It is equal to the kinetic energy of the lighter block.

**Part B**

A. one quarter as fast

B. half as fast

C. the same speed

D. twice as fast

E. four times as fast

We already solved for this in part A. The answer is:

D. twice as fast

Part C

A. The heavy block must be pushed 16 times farther than the light block.

B. The heavy block must be pushed 4 times farther than the light block.

C. The heavy block must be pushed 2 times farther than the light block.

D. The heavy block must be pushed the same distance as the light block.

E. The heavy block must be pushed half as far as the light block.

We can solve for d using the kinematic equation V_{f}^{2} = V_{0}^{2} + 2a(d)

From part A, we know that block 1 has an acceleration 4 times greater than block 2. So if we designate ‘a’ as for block one, the acceleration for block 2 is 0.25a. Assume both blocks start from zero and set up in terms of d:

Block 1:

V_{f}^{2} = V_{0}^{2} + 2a(d)

V_{f}^{2} = 0 + 2a(d)

V_{f}^{2} = 2a(d)

d_{1} = V_{f}^{2} / (2a)

Block 2:

V_{f}^{2} = 2(0.25a)(d)

V_{f}^{2} = 2(0.25a)(d)

V_{f}^{2} = 2(0.25a)(d)

d_{2} = V_{f}^{2} / (2(0.25a))

Set up a ratio:

d_{1} = V_{f}^{2} / (2a)

d_{2} = V_{f}^{2} / (2(0.25a))

So:

= [V_{f}^{2} / (2a)] / [V_{f}^{2} / (2(0.25a))]

= [V_{f}^{2} / (2a)] * [(2(0.25a)) / V_{f}^{2}]

= [(2(0.25a)) * V_{f}^{2}] / [(2a) * V_{f}^{2}]

= [0.25V_{f}^{2}] / V_{f}^{2}]

= 0.25 / 1

= 1 : 4

So:

B. The heavy block must be pushed 4 times farther than the light block.