Answer to Question #180588 in Physics for ankit

Question

A box of mass 15 kg is sliding along a smooth floor with a velocity of 1ms0.15. It then enters a rough portion which has a length of 6.0 m. In this portion of the floor,a frictional force of 80.0 N acts on the box. Determine (i) the work done by the frictional force on the box? (ii) the velocity of the box when it leaves the rough surface and (iii) the length of the rough surface required to bring the box completely to rest.

Answer 1

(i) The work done by friction:

$$W_f=fd=80·6=480\text{ J}.$$

(ii) The velocity when it leaves can be found from conservation of energy (the initial velocity is 15 m/s):

$$\frac12mv_1^2=W+\frac12 mv_2^2,\\\space\\ v_2=\sqrt{v_1^2-\frac{2W}{m}}=12.7\text{ m/s}.$$

(iii) Find the length from conservation of energy:

$$\frac12 mv_1^2=fd,\\\space\\ d=\frac{mv_1^2}{f}=42.2\text{ m}.$$