AP EAMCET · PHYSICS · Work Power Energy
A disc of mass \(100 \mathrm{~g}\) slides down from rest on an inclined plane of \(30^{\circ}\) and come to rest after travelling a distance of \(1 \mathrm{~m}\) along the horizontal plane. If the coefficient of friction is 0.2 for both inclined and horizontal planes, then the work done by the frictional force over the whole journey, approximately, is (Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-1}\) )
- A 0.106 J
- B 0.05 J
- C 0.306 J
- D 0.2 J
Answer & Solution
Correct Answer
(C) 0.306 J
Step-by-step Solution
Detailed explanation
According to the question, acceleration of a block sliding down an inclined plane is shown in the following figure. From third equation of the motion, velocity of disc when it leaves the inclined plane, \( v^2=u^2-2 a s \) or \( u=\sqrt{2 a s} \) \((\because v=0)\) \(\because\)…
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