Question

The displacement x of a particle moving in one dimension, under the action of a constant force is related to time t by the equation $t = \sqrt x + 3$ where x is in meter and t in sec. calculate the work done by the force in the first $6s$ in Joule:
(A) $10$
(B) $0$
(C) $5$
(D) $2$

Answer 1

Hint : to solve this question we have to know about velocity and work. We know that Velocity is a physical vector quantity; both magnitude and direction are needed to define it. The scalar absolute value ( magnitude) of velocity is called speed, being a coherent derived unit whose quantity is measured in the SI ( metric system) as meters per second (m/s)

Complete Step By Step Answer:
we can write the given equation like
$x = {(t - 3)^2}$ According to the question.
We know that velocity
$v = \dfrac{{dx}}{{dt}} = 2(t - 3)$ (After derivation)
${v_f} = 2(6 - 3) = 6$
${v_i} = 2(0 - 3) = - 6$
So, from work- energy theorem
$W = \Delta KE = \dfrac{1}{2}m[{v_f}^2 - {v_i}^2] = \dfrac{1}{2}m[{6^2} - {( - 6)^2}] = 0$
That means the work done by the force in the first six sec is zero. So the right option will be option number B.

Note :
We have to know that work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement. A force is said to do positive work if (when applied) it has a component in the direction of the displacement of the point of application. We can say, A force does negative work if it has a component opposite to the direction of the displacement at the point of application of the force. For example, when a ball is held above the ground and then dropped, the work done by the gravitational force on the ball as it falls is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement).