TS EAMCET · Physics · Work Power Energy
A small body of mass \(500 \mathrm{~g}\) moves on a rough horizontal surface before finally stops. The initial velocity of the body is \(2 \mathrm{~m} / \mathrm{s}\) and coefficient of friction is 0.3 . Then, find absolute value of the average power developed by the frictional force during the time of motion. (Talse, \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(1 \mathrm{~W}\)
- B \(1.5 \mathrm{~W}\)
- C \(2 \mathrm{~W}\)
- D \(2.5 \mathrm{~W}\)
Answer & Solution
Correct Answer
(B) \(1.5 \mathrm{~W}\)
Step-by-step Solution
Detailed explanation
As, retardation is constant, so average velocity \[ v_{a v}=\frac{\mathrm{k}+v}{2}=\frac{2+0}{2}=1 \mathrm{~m} / \mathrm{s} \] Now, friction force, \(f=\mu N=\mu m g\) Given, coefficient of friction, \(\mu=0.3\) Mass of body, \(m=500 \mathrm{~g}=0.5 \mathrm{~kg}\)…
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