First slide

Work Energy Theorem

Question

A block of mass 2 kg rests on ground. An external agent raises the block to a height of 5m above ground where its speed is V. It work done by the external agent is 200 J, value of V is

Moderate

A

10 m/s
B

5 m/s
C

4 m/s
D

20 m/s

Solution

Work done by the external agent
= Change in kinetic energy + change in P.E.
$\begin{array}{l} ∴ 200 = \frac{1}{2} {mV}^{2} + mgh \\ \Rightarrow 200 = \frac{1}{2} \times 2 \times V^{2} + 2 \times 10 \times 5 \Rightarrow V = 10 m / s \end{array}$

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