Answer to Question #102689 in Mechanics | Relativity for sakshi

Question #102689
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
Expert's answer

(i) The law of conservation of the energy states




"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."

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