The distance $x$ moved by a body of mass 0.5 kg due to a force varies with time $t$ as
$x = 3 t^{2} + 4 t + 5$
where $x$ is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?

25 J

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50 J

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75 J

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100 J

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Solution

The correct option is C
75 J

Velocity, $v = \frac{d x}{d t} = \frac{d}{d t} (3 t^{2} + 4 t + 5) = 6 t + 4$
Acceleration, $a = \frac{d v}{d t} = \frac{d}{d t} (6 t + 4) = 6 m s^{- 2}$
Therefore, applied force, $F = m a = 0.5 \times 6 = 3 N$
Now, the distance moved in 2 s, $x = 3 \times (2)^{2} + (4 \times 2) + 5 = 25 m$
$∴$ Work done, $W = F x = 3 \times 25 = 75 J$
Hence, the correct choice is (c).

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The distance $x$ moved by a body of mass 0.5 kg due to a force varies with time $t$ as
$x = 3 t^{2} + 4 t + 5$
where $x$ is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?

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The distance x moved by a body of mass 0.5 kg due to a force varies with time t as x=3t2+4t+5 where x is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?

The correct option is C 75 J Velocity, v=dxdt=ddt(3t2+4t+5)=6t+4 Acceleration, a=dvdt=ddt(6t+4)=6 ms−2 Therefore, applied force, F=ma=0.5×6=3N Now, the distance moved in 2 s, x=3×(2)2+(4×2)+5=25 m ∴ Work done,W=Fx=3×25=75J Hence, the correct choice is (c).

The distance $x$ moved by a body of mass 0.5 kg due to a force varies with time $t$ as
$x = 3 t^{2} + 4 t + 5$
where $x$ is expressed in metre and t in second. What is the work done by the force in the first 2 seconds?