Question

A 4.0 kg crate has initial speed 2.70 m/sec and slides 1.80 meters across a flat,

rough floor before coming to rest.

a) Find the initial kinetic energy and final kinetic energy.

b) How much mechanical energy is lost due to friction?

c) What work do the following forces do?

a. The gravitational force

b. The normal force

c. Friction

Answer 1

(a)

$$KE_i=\dfrac{1}{2}mv_i^2=\dfrac{1}{2}\cdot4.0\ kg\cdot(2.7\ \dfrac{m}{s})^2=14.58\ J,$$ $$KE_f=\dfrac{1}{2}mv_f^2=\dfrac{1}{2}\cdot4.0\ kg\cdot(0\ \dfrac{m}{s})^2=0\ J.$$

(b) By the law of conservation of mechanical energy, we have:

$$KE_i+PE_i+W_{ext}=KE_f+PE_f,$$ $$KE_i+0+W_{ext}=0,$$ $$W_{ext}=-KE_i=-14.58\ J.$$

Therefore, 14.58 J of mechanical energy is lost due to friction.

(c) a. In this case the gravitational force acts downward on the crate as it displaced rightward. Therefore, the force vector and the displacement vector are at right angles to each other. As a result, the angle between $F_g$ and $d$ is 90 degrees. Therefore, the work done by the gravitational force equals zero:

$$W_g=F_gdcos90^{\circ}=0.$$

b. In this case the normal force acts upward on the crate as it displaced rightward. Therefore, the force vector and the displacement vector are at right angles to each other. As a result, the angle between $F_N$ and $d$ is 90 degrees. Therefore, the work done by the normal force equals zero:

$$W_N=F_Ndcos90^{\circ}=0.$$

c. In this case the force of friction acts leftward on the crate as it displaced rightward. Therefore, the force vector and the displacement vector are in the opposite direction. As a result, the angle between $F_{fr}$ and $d$ is 180 degrees. Therefore, the work done by the normal force equals:

$$W_{fr}=F_{fr}dcos180^{\circ}=\Delta KE=-KE_i=-14.58\ J.$$