First slide

Work Energy Theorem

Question

You crouch from standing position, lower your centre of mass by 20 cm and then jump vertically upward into the air. The average force that the ground exerts on you while you are jumping is three times your weight. Then the speed with which you leave the ground in standing position is

Easy

A

7 m/s
B

2.8 m/s
C

3.6 m/s
D

2.2 m/s

Solution

(2)
$\begin{array}{l} F_{n e t} = N e t v e r t i c a l l y u p w a r d f o r c e o n m e \\ = N - m g = 3 m g - m g = 2 m g \\ B y w o r k - e n e r g y T h e o r e m, \frac{1}{2} m v^{2} - 0 = F_{n e t} \times h \\ \Rightarrow \frac{1}{2} m v^{2} = 2 m g \times h \\ \Rightarrow v = \sqrt{4 g h} = \sqrt{4 \times 10 \times 0.2} m / s = 2 \sqrt{2} m / s \\ \Rightarrow v = 2 \times 1.414 m / s = 2.828 m / s \end{array}$

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