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Work Power and Energy question

2006 · Shift 0 · Q124

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Work Power and Energy
2006 · Shift 0 · Q124

Work Power and Energy question

2006 · Shift 0 · Q124

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1

A ball of mass

0.2kg

is thrown vertically upwards by applying a force by hand. If the hand moves

0.2m

while applying the force and the ball goes upto

2m

height further, find the magnitude of the force. (consider

g = 10\,m/{s^2}

).

A
$4N$
B
$16N$
C
$20N$
D
$22N$

View written solutionFree

Correct answer: D

Given data

Mass of ball: $m = 0.2\,\text{kg}$
Distance moved by hand while applying force: $s_1 = 0.2\,\text{m}$
Additional height reached by ball after leaving hand: $s_2 = 2\,\text{m}$
Acceleration due to gravity: $g = 10\,\text{m/s}^2$

We need to find the force $F$ applied by the hand.

Find the speed of the ball at the instant it leaves the hand

After leaving the hand, the ball rises through height $2\,\text{m}$ and finally comes to rest at the top.

Using $v^2 = u^2 - 2gs$ At the highest point, $v=0$ , so $0 = u^2 - 2g(2)$ $u^2 = 2 \cdot 10 \cdot 2 = 40$

So the kinetic energy of the ball at release is $\frac{1}{2}mu^2 = \frac{1}{2}(0.2)(40) = 4\,\text{J}$

Apply work-energy theorem during contact with the hand

While the hand is in contact, the ball moves upward by $0.2\,\text{m}$ .

Forces on the ball during this motion:

Upward applied force = $F$
Downward weight = $mg = 0.2 \cdot 10 = 2\,\text{N}$

Work done by hand: $W_F = F(0.2)$

Work done by gravity: $W_g = -mg(0.2) = -2(0.2) = -0.4\,\text{J}$

Net work done equals gain in kinetic energy: $F(0.2) - 0.4 = 4$ $0.2F = 4.4$ $F = \frac{4.4}{0.2} = 22\,\text{N}$

Check options

The correct option is: $\boxed{22\,\text{N}}$

So, Option D is correct.

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