<p>A solid sphere of weight 40N rolls smoothly up an incline at an angle of 30 d

Question

<p>A solid sphere of weight 40N rolls smoothly up an incline at an angle of 30 degrees. At the bottom of the incline the center of mass of the sphere has a translational speed of 5m/s.</p>
<p>A) What is the kinetic energy of the sphere at the bottom of the incline?</p>
<p>B) What is the speed of the center of mass once the sphere has rolled up along the incline a distance of 1m?</p>
<p>C) How far does the sphere travel up the incline (as measured along the surface of the incline)?</p>
<p>D) What is the acceleration of the center of mass as the sphere rolls up the incline?</p>

Explanation / Answer

a)kinetic energy=1/2 m v^2 = 0.5 *4* 25=50 J b)increase in potential energy= 4*g*1=40 J total energy=50 J ==>kinetic energy @1m frm bottom=10 J =>1/2*4*v^2= 10 ==>v=sqrt(5) c)travels till it attains poten energy=50 J ==>4*g*h = 50 J ==>h=1.25m.

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