Work Energy Power

JEE Main level, numerical answer, medium

151 of 183

Question 151 of 183

A block of mass 5kg5kg starting from rest pulled up on a smooth incline plane making an angle of 30∘30^{\circ} with horizontal with an effective acceleration of 1ms−21m{s}^{-2}. The power delivered by the puling force at t=10st=10s from the start is _____ WW.
[Use
g=10ms−2g=10m{s}^{-2}]
(Calculate the nearest integer value)

Show answer and solution

Answer: 300

Let's consider the following free-body diagram:
From the above diagram, it can be written that

F−mgsin⁡30∘=ma...(1)F-mg\sin 30^{\circ}=ma...\left(1\right)
Substitute the values of the known parameters into equation (1) and solve to calculate the force applied on the block.

F−50sin⁡30∘=5×1⇒F=(5+25)N=30N\begin{matrix}F-50\sin 30^{\circ} & = & 5\times 1 \\ & \Rightarrow & F=\left(5+25\right)N \\ & = & 30N\end{matrix}
The velocity of the block
(v)\left(v\right) at t=10st=10s can be written as
v=0+1ms−2×10s=10ms−1\begin{matrix}v & = & 0+1m{s}^{-2}\times 10s \\ & = & 10m{s}^{-1}\end{matrix}
Hence, the required power is given by

P=Fv=30N×10ms−1=300W\begin{matrix}P & = & Fv \\ & = & 30N\times 10m{s}^{-1} \\ & = & 300W\end{matrix}

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