Question

A force $F = (3t\hat i + 5\hat j)N$ acts on a body due to which its displacement varies as $S = (2{t^2}\hat i - 5\hat j)m$ . Work done by this force in $2s$ is
$(A)32J$
$\left( B \right)24J$
$(C)46J$
$(D)20J$

Answer 1

The work done by a force is the product of the force and the change in displacement. For a non-uniform field, we can calculate work done for a small portion and then these small amounts are added through the integration method taking a limit of time. Here in the problem, the change in displacement has to be calculated w.r.t the time since the given displacement is a function of time and thereafter, this is to multiply with the given force quantity. To find the work done by the force within the given time the integration has to be done with the time limit.

Formula used:
The work done by the force $F$ is, $W = \int_0^t {F.dS}$
$ds =$ change in displacement
$t =$ time.

Complete answer:
The force is given by, $F = (3t\hat i + 5\hat j)N$
The body is displaced due to the applied force as $S = (2{t^2}\hat i - 5\hat j)m$
Now, if we derive the displacement w.r.t the time the result will be, $\dfrac{{dS}}{{dt}} = 4t\hat i$
So, $dS = 4t\hat i.dt$
The work done by a force is the product of the force and the change in displacement.
Hence, the work done by the force $F$ is, $W = \int_0^t {F.dS}$
$\Rightarrow W = \int_0^t {(3{t^2}\hat i + 5\hat j).4t\hat i} dt$
$\Rightarrow W = \int_0^2 {12{t^2}dt}$
$\Rightarrow W = \left[ {\dfrac{{12{t^3}}}{3}} \right]_0^2$
$\Rightarrow W = 4({2^3} - {0^3})$
$\Rightarrow W = 32$
So, the work done by the force $32{\text{Joule}}$
Hence the right answer is in option $(A) \Rightarrow 32J$

Note:
The work is done when energy is passed from one place to another. The Work is done when a force causes a body to move. When work is done against frictional forces applying to a body, the body’s temperature increases.
One joule of work is done when one newton of force results in a displacement of one meter. This implies that work done is also measured in newton meters ($Nm$).
$1J = 1Nm$
It is important not to confuse newton-meters with Newton-meters (used to measure weight in spring balances).