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Work Power and Energy question

2022 · 25 Jul · Shift 1 · Q44

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Work Power and Energy
2022 · 25 Jul · Shift 1 · Q44

Work Power and Energy question

2022 · 25 Jul · Shift 1 · Q44

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1

A body of mass

0.5 \mathrm{~kg}

travels on straight line path with velocity

v=\left(3 x^{2}+4\right) \mathrm{m} / \mathrm{s}

. The net workdone by the force during its displacement from

x=0

to

x=2 \mathrm{~m}

is :

A
64 J
B
60 J
C
120 J
D
128 J

View written solutionFree

Correct answer: B

Use work-energy theorem

The net work done by the force is equal to the change in kinetic energy:

$W_{\text{net}} = \Delta K = \frac{1}{2}m\left(v_f^2 - v_i^2\right)$

Given:

Mass, $m = 0.5\,\text{kg}$
Velocity as a function of position: $v = 3x^2 + 4$

We need velocity at:

$x=0$
$x=2$

Find initial velocity

At $x=0$ :

$v_i = 3(0)^2 + 4 = 4\,\text{m/s}$

Find final velocity

At $x=2$ :

$v_f = 3(2)^2 + 4 = 3\cdot 4 + 4 = 16\,\text{m/s}$

Compute change in kinetic energy

$W_{\text{net}} = \frac{1}{2}(0.5)\left(16^2 - 4^2\right)$

$= 0.25(256 - 16)$

$= 0.25 \times 240$

$= 60\,\text{J}$

Match with options

$60\,\text{J}$ corresponds to Option B.

Comparison with stored answer

Stored correct answer is B, which matches our derived answer.

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