Question

A block of mass m = 1 kg, moving on a horizontal surface with speed vi = 2 ms–1 enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force Fr on the block in this range is inversely proportional to x over this range, Fr = -k/x for 0.1 < x < 2.01 m = 0 for x < 0.1m and x > 2.01 m where k = 0.5 J. What is the final kinetic energy and speed vf of the block as it crosses this patch?

Answer 1

The change of energy is equal to work done, so

$$\frac{mv_f^2}{2}-\frac{mv_i^2}{2}=W$$

Thus

$$v_f=\sqrt{\frac{2W}{m}+v_i^2}$$

The work done is given by

$$W=\int_{x_1}^{x_2}Fdx=-k\int_{x_1}^{x_2}\frac{dx}{x}$$

$$=-k\ln\frac{x_2}{x_1}=-0.5\ln\frac{2.01}{0.1}=-1.5\:\rm J$$

Therefore

$$v_f=\sqrt{\frac{2\times(-1.5)}{1}+2^2}=1\:\rm m/s$$

The final kitetic energy

$$K_f=\frac{mv_f^2}{2}=\frac{1\times 1^2}{2}=0.5\:\rm J$$