Question

An object at a vertical elevation of 20 m and a speed of 5 m/s decreases in elevation to an elevation of 1 m. At this location, the object has a velocity of 15 m/s. The mass of the object is 68 kg. Assuming the object is the system, determine if there is any work transfer associated with the object (there is no heat transfer). The object is solid, incompressible and its temperature does not change during the process. If there is work transfer, is work done on or by the object? Assume the acceleration of gravity g = 9.81 m/s2.

Answer 1

Answer with Explanation:

We know that from the principle of work and energy we have

Work done on/by a body =ΔEnergy of the body.

Now as we know that energy of a body is the sum of it's kinetic and potential energy hence we can find out the magnitude of the final and initial energies as explained under

$E_{initial}=P.E+K.E\\\\E_{initail}=mgh_1+\frac{1}{2}mv_1^{2}\\\\68\times 9.81\times 20+\frac{1}{2}\times 68\times (5)^{2}=14191.6Joules$

Similarly the final energy is calculated to be

$E_{final}=P.E+K.E\\\\E_{final}=mgh_2+\frac{1}{2}mv_2^{2}\\\\68\times 9.81\times 1+\frac{1}{2}\times 68\times (15)^{2}=8317.08Joules$

As we can see that the energy of the object has changed thus by work energy theorem we conclude that work transfer is associated with the object.

Part 2)

The change in the energy of the body equals $8317.08-14191.6=-5874.52Joules$

Since the energy is lost by the system hence we conclude that work is done by the object.