A small block with mass 0.0500 kg slides in a vertical circle of radius 0.425 m

Question

A small block with mass 0.0500 kg slides in a vertical circle of radius 0.425 m on the inside of a circular track. During one of the revolutions of the block, when the block is at the bottom of its path, point A, the magnitude of the normal force exerted on the block by the track has magnitude 3.75 N . In this same revolution, when the block reaches the top of its path, point B, the magnitude of the normal force exerted on the block has magnitude 0.685 N .

How much work was done on the block by friction during the motion of the block from point A to point B?

Explanation / Answer

At the bottom

3.75 =0.0500*9.81+0.0500*v2/0.425

v2 =(3.2595)(0.425)/0.0500

Then speed v =5.263m/s

At the top,

0.685 =0.0500*u2/0.425-0.0500*9.81

0.685 =0.0500*u2/0.425-0.4905

u =3.160m/s

From above equations we can write it as How much work was done on the block by friction during the motion of the block from point A to point B

(1/2)m(u2-v2)=-mg(2R)+Wf

Wf =mg(2R)+(1/2)m(u2-v2) =0.0500*9.81*2*0.425+0.5*0.0500(9.991-27.699)=0.4169+0.025(-17.708)=-0.0258Joules

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