Question

The distance X moved by a body of mass 0.5 kg
under the action of a force varies with time t as
x(m) = 3t2 + 4t + 5 (here t is expresed in second).
What is the work done by the force in first
2 seconds?
(1) 20J
(2) 40 J
(3) 50 J
(4) 60 J​

Answer 1

Answer:

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Answer 2

Answer:

Work done will be 60joule

Explanation:

In this question,

We have been given that

Mass of the body is 0.5k.g

Force varies with time i.e $x(m)=3t^{2}+4t+5$

We need to find Work done by the force in first 2 seconds

According to the question,

$x(m)=3t^{2}+4t+5$

Differentiating with respect to t we get,

$\frac{\mathrm{d} x}{\mathrm{d} t}=6t+4$

Again differentiating with respect to t we get,

$\frac{\mathrm{d}^2 x}{\mathrm{d} t^2}=6$

$[\frac{\mathrm{d}^2 x}{\mathrm{d} t^2}]_{t=2}=6$

Therefore Acceleration=6$m/s^{2}$

Now calculating distance we get,

D=$[x]_{t=2} - [x]_{t=0}$

D=$3(2)^{2}+4\times 2+5 - 5$

D=12+8

D=20m

Work done= Force×Distance

As force is given by Ma i.e mass×Acceleration

Work done=Ma×Distance

Now putting the values we get,

Work done=0.5×6×20

Work done=60 joule

So, Work done by the force is equal to 60 joule