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Work Power and Energy question

2016 · 10 Apr · Shift 1 · Q46

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Work Power and Energy
2016 · 10 Apr · Shift 1 · Q46

Work Power and Energy question

2016 · 10 Apr · Shift 1 · Q46

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1

Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on the body in first two seconds of the motion is : JEE Main 2016 (Online) 10th April Morning Slot Physics - Work Power & Energy Question 107 English

JEE Main 2016 (Online) 10th April Morning Slot Physics - Work Power & Energy Question 107 English

A
12000 J
B
$-$ 12000 J
C
$-$ 4500 J
D
$-$ 9300 J

View written solutionFree

Correct answer: C

Use work–energy theorem

The net work done on the body in a time interval is equal to the change in its kinetic energy:

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

Here, mass of the body is

$m = 10\,\text{kg}$

Read velocities from the velocity–time graph

From the graph, during the first two seconds:

initial velocity at $t=0$ is $v_i = 30\,\text{m/s}$
final velocity at $t=2\,\text{s}$ is $v_f = 0\,\text{m/s}$

Calculate change in kinetic energy

$W = \frac{1}{2}(10)\left(0^2 - 30^2\right)$

$W = 5(0 - 900)$

$W = -4500\,\text{J}$

Match with the options

Thus, the work done on the body in the first two seconds is

$\boxed{-4500\,\text{J}}$

So the correct option is:

C: $-4500\,\text{J}$

Compare with stored correct answer

Stored correct answer is C.

My derived answer is also C, so they agree.

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