Answer to Question #86657 in Mechanics | Relativity for Ogooluwa

Question

A body of mass 2500g is projected up an inclined plane with initial velocity 300cm3^-1. The plane is inclined at an angle 15° to the horizontal, and the coefficient of friction between the plane and the body is 0.25. determine how far up the plane it will go before coming to rest.

Answer 1

The Newton's second law states

$ma=-mg\sin\theta-\mu mg\cos\theta$

The acceleration of a body

$a=-g(\sin\theta+\mu \cos\theta)=-9.8(\sin 15^{\circ}+0.25\cos 15^{\circ})$

$=-4.9\:\rm{m/s^2}$

If the initial velocity is 10 m/s, the distance traveled by body before it will coming to rest

$d=\frac{-v_i^2}{2a}=\frac{-(10\:\rm{m/s})^2}{2\times (-4.9)\:\rm{m/s^2}}=10.2\:\rm{m}$