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.

Respuesta :

Answer:1.63 m

Explanation:

Given

mass of block [tex]m=0.5 kg[/tex]

inclination [tex]\theta =30^{\circ}[/tex]

Amount of work done [tex]W=4 J[/tex]

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is [tex]s\sin \theta [/tex]

Change in Potential Energy [tex]=mg(\sin \theta -0)[/tex]

[tex]\Delta P.E.=0.5\times 9.8\times s\sin 30[/tex]

[tex]4=0.5\times 9.8\times s\sin 30[/tex]

[tex]s=\frac{4}{2.45}[/tex]      

[tex]s=1.63 m[/tex]