Question

A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of work and the block slides a distance s along the incline before it stops. Determine the value of s.

Answer 1

Answer:1.63 m

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30^{\circ}$

Amount of work done $W=4 J$

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is $s\sin \theta$

Change in Potential Energy $=mg(\sin \theta -0)$

$\Delta P.E.=0.5\times 9.8\times s\sin 30$

$4=0.5\times 9.8\times s\sin 30$

$s=\frac{4}{2.45}$      

$s=1.63 m$