Question

Under the action of a force, a 2 kg body moves
- such that its position x as a function of time t is
given by x= t^2/3, where x is in metre and t in
second. The work done by the force in first two
seconds is
(1) 1600 J
(2) 160
(3) 160
(4) 16/9

Answer 1

Answer:

Work done by the force is $\dfrac{16}{9}\ Joules$.

Explanation:

It is given that,

Mass of the object, m = 2 kg

Its position as a function of time t is given by :

$x=\dfrac{t^2}{3}$

Velocity, $v=\dfrac{dx}{dt}=\dfrac{2t}{3}$

Where

x is in meters

t is in seconds

We need to find the work done by the force in first two seconds. The work done by an object is given by :

$W=\dfrac{1}{2}m(v^2-u^2)$................(1)

Where

v is the final velocity

u is the initial velocity

At t = 0, u = 0

At t = 2 s, $v=\dfrac{4}{3}$

Put the values of u and v in equation (1), we get :

$W=\dfrac{1}{2}\times 2((4/3)^2-0)$

$W=\dfrac{16}{9}\ J$

So, the work done by the force in first two  seconds is $\dfrac{16}{9}\ Joules$. Hence, this is the required solution.

Answer 2

Explanation:

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