Question

A force acts on a 2 Kg object so that its position is given as a function of times as $x = 3{t^2} + 5$. What is the work done by this force in the first 5 seconds?

A. 850 J

B. 900 J

C. 950 J

D. 875 J

Answer 1

Hint: The work done equals change in kinetic energy. Here, initial kinetic energy is zero, therefore, ${\text{W}} = \dfrac{1}{2}{\text{m}}{{\text{v}}^2}$. Here M=2 Kg. Velocity equals differentiation of displacement with respect to time. We use the given equation to find the velocity at t=5 seconds.

Formula used:

$F=ma$

Here, $m$ is the mass of the body and $a$ is the acceleration of the body.

Complete step-by-step answer:

Given mass of an object = 2Kg

Given function of time: $x = 3{t^2} + 5$

Force: A force is a push or pull upon an object resulting from the object's interaction with another object.

Newton’s second law, F= ma

Where F is force, m is mass of an object and ‘a’ is acceleration.

Differentiating the above equation with respect to time gives velocity

$ \Rightarrow $${\text{v = }}\dfrac{{dx}}{{dt}}$

Differentiating $x = 3{t^2} + 5$ on both sides with t

$\dfrac{{dx}}{{dt}} = 6t + 0$

$v = 6t + 0$

At ${\text{t = 0 }} \Rightarrow {\text{v = 0}}$

${\text{t = 5 sec }} \Rightarrow {\text{ v = 30 m/s}}$

When a force causes a body to move, work is being done on the object by the force. Work done is the change in kinetic energy when a force (F) moves an object through a distance (d). 

so: energy transferred= work done.

The S.I unit of work is joule (J).

Work done$ = \Delta K.E$

Where KE is the Kinetic Energy.

Work done$ = \dfrac{1}{2}{\text{m}}{{\text{v}}^2} - 0$

$ = \dfrac{1}{2}\left( 2 \right){\left( {30} \right)^2}$

= 900 J

Therefore, correct answer is option (B).

Note: The work done by a force on an object can be positive, negative, or zero, depending upon the direction of displacement of the object with respect to the force. For an object moving in the opposite direction to the direction of force, such as friction acting on an object moving in the forward direction, the work done due to friction of force is negative.