A 3.0-kg mass is dropped from the edge of a 50-m tall building with an initial speed of zero. The mass strikes the ground with a downward velocity of 25 m/s. Find the change in mechanical energy of the mass caused by air resistance between the point where it is dropped and the point where it strikes the ground?
A) −0.46 kJ
B) −0.53 kJ
C) −0.61 kJ
D) −0.38 kJ
E) −0.81 kJ

Question

Quizplus · Accepted Answer

The change in mechanical energy is equal to the work done by non-conservative forces, such as air resistance. We can calculate the gravitational potential energy the mass has at the initial height using:

$U_{i} = mgh = (3.0 kg)(9.81 m/s^2)(50 m) = 1471.5 J$

We can also calculate the kinetic energy the mass has just before it hits the ground using:

$K_{f} = \frac{1}{2}mv_{f}^2 = \frac{1}{2}(3.0 kg)(25 m/s)^2 = 937.5 J$

The total mechanical energy at the initial height is therefore:

$E_{i} = U_{i} = 1471.5 J$

The total mechanical energy just before the mass hits the ground is:

$E_{f} = U_{f} + K_{f}$

At this point, the mass is at ground level so its gravitational potential energy is zero:

$E_{f} = K_{f} = 937.5 J$

Therefore, the change in mechanical energy is:

$\Delta E = E_{f} - E_{i} = K_{f} - U_{i} = 937.5 J - 1471.5 J = -534 J = -0.53 kJ$

So, the answer is B.

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