Question

A particle P of mass 1.2kg is released from rest at the top of a slope and start to move.The slope as length 4m and is inclined at 25degree to the horizontal.The coefficient of friction between P and the slope is 0.25. find (i) the frictional component of the contact force on p (ii) the acceleration (iii) the speed with which P reaches the buttom of the slope.( Take g=10m/s)

Answer 1

(i) The frictional component of the contact force on P

$$F_{\rm fr}=\mu N=\mu mg\cos\theta$$

$$=0.25\times 1.2\times 10\times \cos 25^{\circ}=2.72\: \rm N$$

(ii) The acceleration

$$a=\frac{mg\sin\theta-F_{\rm fr}}{m}$$

$$=\frac{1.2\times 10\times\sin 25^{\circ}-2.72}{1.2}=1.96\:\rm m/s^2$$

(iii) The speed with which P reaches the bottom of the slope

$$v=\sqrt{2ad}=\sqrt{2\times 1.96\times 4}=3.96\:\rm m/s$$