Question

A worker pushes a wood crate of mass 14 kg across a concrete floor with a constant horizontal force of magnitude 40N. In a straight-line displacement of magnitude d= 0.50 m, the speed of the crate decreases from 0.60 m/s to 0.20 m/s.

Find the following:

a. Work done by the worker’s force.

b. Change in the kinetic energy of the crate.

c. Increase the thermal energy of the crate and the floor.

Answer 1

a) $W = 40N\cdot 0.5m = 20J$

b) $\Delta K = \dfrac{mv_f^2}{2}-\dfrac{mv_i^2}{2} = \dfrac{14\cdot 0.2^2}{2}-\dfrac{14\cdot 0.6^2}{2}\approx-2.2J$

c) According to the energy-work theorem, the increase of the thermal energy is:

$$E_t = -\Delta K + W =2.2J + 20J \approx 22J$$