You hold an object weighing 10 N so that it is at rest.
How much work do you do on the object? How much work does gravity do on the object?

The work you do is:

1. zero
2. positive
3. negative

Both you and gravity do zero work, because the displacement is zero.

You raise the 10 N object 0.5 m vertically. The object starts and ends at rest.

How much work do you do on the object?

1. 5 J
2. More than 5 J
3. Less than 5 J

Both the initial and final kinetic energy is zero. The change in potential energy is 5 J, and that comes from the work you do. You do 5 J worth of work.

You move the 10 N object 2 m horizontally. The object starts and ends with the same speed.

How much work does gravity do on the object? How much work do you do on the object?

The work you do on the object is:

1. 0
2. 20 J
3. some positive value
4. some negative value

Your upward force and the downward force of gravity are perpendicular to the displacement, so neither force does any work.

A car traveling at an initial speed v on a flat road comes to a stop in a distance L once the brakes are applied. Friction is the force that brings the car to rest. Assuming the same force is applied, how far does the car travel once the brakes are applied if the initial speed is 2v?

1. L x 2^½
2. 2L
3. 3L
4. 4L

The answer is 4L. The kinetic energy has increased by a factor of 4, so the work required to bring the car to rest must increase by a factor of 4. If the force is unchanged the distance must increase by a factor of 4.

A box with a mass of 1.0 kg is moving to the right. Its initial speed is 2 m/s. After traveling 3 m, its speed is 4 m/s. In addition to the usual forces (gravity, normal, friction) there are two other forces acting on the box. One is a 5 N force that has an upward vertical component of 4 N and a component to the left of 3 N. The other force is 8 N to the right.

(a) What is the normal force?

(b) What is the coefficient of kinetic friction?

A good free-body diagram is needed here.

Applying Newton's second law in the y-direction, it can be shown that the normal force is N = 9.8 - 4 = 5.8 N.

Applying Newton's second law in the x-direction tells us:

ΣF_x = 8 - 3 - f_k = 5 - f_k = ma_x

Finding μ_k takes more work. Let's try the energy method.

Energy method

Start with the master energy equation:

U_i + K_i + W_nc = U_f + K_f

There is no change in potential energy, so we can get rid of the potential energy terms. This gives:

K_i + W_nc = K_f

W_nc = K_f - K_i = ½ mv_f² - ½ mv_o²

W_nc = 8 J - 2 J = 6 J

In this case W_nc = 6 = (ΣF_x) x (3 m), so:

ΣF_x = 6/3 = 2 N

We showed already that ΣF_x = 5 - f_k, so:

f_k = 5 - ΣF_x = 5 - 2 = 3 N

We found the normal force to be 5.8 N, so:

μ_k = f_k / N = 3/5.8 = 0.52

Could we have done this problem using constant acceleration equations? Absolutely. We would have found the acceleration from the equation:

v_x ² = v_ox² + 2 a_x (x - x_o)

If we multiply through by ½ m, the equation becomes:

½ mv_x ² = ½ mv_ox² + ma_x (x - x_o)

This says K_f = K_i + F_x Δx, which is our master energy equation without the potential energy terms.

That constant acceleration equation is actually an energy equation in disguise - so we've been using energy concepts all along.

Three identical balls are launched with the same initial speed from the top of a cliff overlooking flat ground. Ball A is launched horizontally. Ball B has an initial velocity directed 20 degrees below the horizontal. Ball C has an initial velocity directed 40 degrees above the horizontal. Which ball hits the ground with the highest speed?

1. Ball A
2. Ball B
3. Ball C
4. Equal for all three

You could try to analyze each ball separately using the projectile motion equations, but that would be much more difficult than applying conservation of energy ideas:

U_i + K_i = U_f + K_f

The initial and final potential energies are the same for each ball, because they all start from the same height and end at the same lower height at ground level. They all have the same initial kinetic energies...so, they all must have the same final kinetic energies.

Therefore, each ball hits the ground with the same speed.

Note that it doesn't matter what the initial angle of the ball is. That affects the direction of the velocity each ball has on impact, but not the magnitude (i.e., the speed).