Answer to Question #102689 in Mechanics | Relativity for sakshi

Question

A block of mass 4.0 kg starts from rest and slides down a surface which corresponds to a quarter of a circle of 2.0 m radius. (i) If the curved surface is smooth, find the speed at the bottom. (ii) If the speed at the bottom is 2.0 ms-1, calculate the energy dissipated due to friction in the descent. (iii) After the block reaches the horizontal with a speed of 2.0 ms-1it slides to a stop in a distance of 1.5 m. Calculate the frictional force acting on the horizontal surface. Take g = 10.0m/s²

Answer 1

(i) The law of conservation of the energy states

$$E_i=E_f.$$

Hence

$$mgh=mv_f^2/2,$$

$$v_f=\sqrt{2gh}=\sqrt{2\times 10\times 2.0}=6.3\:\rm m/s.$$

(ii) The change of energy due friction

$$\Delta E=E_i-E_f=mgh-mv_f^2/2$$

$$=4.0\times 10\times 2.0-4.0\times 2.0^2/2=72\:\rm J.$$

(iii) The work done by friction force on the horizontal surface

$$W=Fd=mv_f^2/2-0=4.0\times 2.0^2/2=8.0\:\rm J.$$

Thus, the friction force

$$F=8.0/1.5=5.3\:\rm N.$$