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Work Power and Energy question

2023 · 8 Apr · Shift 2 · Q61

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Work Power and Energy
2023 · 8 Apr · Shift 2 · Q61

Work Power and Energy question

2023 · 8 Apr · Shift 2 · Q61

JEE MainPhysicsWork Power and EnergyNumerical+4 / −1

A body of mass

5 \mathrm{~kg}

is moving with a momentum of

10 \mathrm{~kg} \mathrm{~ms}^{-1}

. Now a force of

2 \mathrm{~N}

acts on the body in the direction of its motion for

5 \mathrm{~s}

. The increase in the Kinetic energy of the body is

\underline{\hspace{2cm}}\mathrm{J}

.

Numerical answer

View written solutionFree

Correct answer: 30

Given data

Mass of body: $m = 5\,\text{kg}$
Initial momentum: $p_i = 10\,\text{kg m s}^{-1}$
Force applied: $F = 2\,\text{N}$
Time of action: $t = 5\,\text{s}$

Find initial velocity

Using momentum, $p = mv$ So, $v_i = \frac{p_i}{m} = \frac{10}{5} = 2\,\text{m/s}$

Find acceleration produced by the force

Using Newton’s second law, $F = ma$ $a = \frac{F}{m} = \frac{2}{5} = 0.4\,\text{m/s}^2$

Find final velocity after 5 s

Using $v_f = v_i + at$ $v_f = 2 + (0.4)(5) = 2 + 2 = 4\,\text{m/s}$

Compute initial kinetic energy

$K_i = \frac{1}{2}mv_i^2$ $K_i = \frac{1}{2}(5)(2^2) = \frac{5}{2}\cdot 4 = 10\,\text{J}$

Compute final kinetic energy

$K_f = \frac{1}{2}mv_f^2$ $K_f = \frac{1}{2}(5)(4^2) = \frac{5}{2}\cdot 16 = 40\,\text{J}$

Increase in kinetic energy

$\Delta K = K_f - K_i = 40 - 10 = 30\,\text{J}$

Therefore, the increase in kinetic energy is: $\boxed{30}$

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