Question 346: Work, Energy and Power & Circular Motion

Question

A small block of mass 1 kg is released from rest at the top of a rough track. The track is circular arc of radius 40 m. The block slides along the track without toppling and a frictional force acts on it in the direction opposite to the instantaneous velocity. The work done in overcoming the friction up to the point Q, as shown in the figure, below, is 150 J. (Take the acceleration due to gravity, g = 10 m/s
2
)
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l723h1dh/6aa63041-1c3b-4f10-9d53-29f1524c45cd/481d1650-20a2-11ed-8b98-7f5599df755c/file-1l723h1di.png?format=png)

Accepted Answer

Here we will use the concept of work-energy theorem. By work-energy theorem,
Total work done = Change in KE
$\Rightarrow$ Workdone by gravity + workdone by friction = $\Delta KE$
$ \Rightarrow mgR\sin 30^\circ  - 150 = {1 \over 2}m{v^2} - 0$
$ \Rightarrow 1 	imes 10 	imes 40 	imes {1 \over 2} - 150 = {1 \over 2}{v^2}$
$ \Rightarrow 200 - 150 = {1 \over 2}{v^2}$
$ \Rightarrow {v^2} = 100$
$ \Rightarrow v = 10$ m/s
Hence, option (B) is correct.
![](https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lzug0j2o/520648c7-3822-49c7-b468-85c70c398a57/5480b900-5a8f-11ef-b2ce-1d34423c56ea/file-jaoe38c1lzug0j2p.png?format=png)

Answer

5 ms
-1

Answer

10 ms
-1

Answer

$10\sqrt 3 $ ms
-1

Answer

20 ms
-1

Work, Energy and Power & Circular Motion

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