Question

A force \[F = \left( {3x + 4} \right)\] newton displaces an object of mass 10 kg from \[x = 0\] to \[x = 4\,{\text{m}}\]. If the speed of the object at \[x = 0\] is \[2\,{\text{m}}{{\text{s}}^{ - 1}}\], its speed at \[x = 4\,{\text{m}}\] will be
A. \[3\,{\text{m}}{{\text{s}}^{ - 1}}\]
B. \[\sqrt 3 \,{\text{m}}{{\text{s}}^{ - 1}}\]
C. \[2\sqrt 3 \,{\text{m}}{{\text{s}}^{ - 1}}\]
D. \[4\,{\text{m}}{{\text{s}}^{ - 1}}\]

Answer 1

Hint: The force acting on the particle is a variable force. Recall the formula for the work done by the variable force and calculate the work done. According to the work-energy theorem, the work done is the change in kinetic energy.

Formula used:
Work done, \[W = \int\limits_{{x_i}}^{{x_f}} {F\left( x \right)\,dx} \]
Here, \[F\left( x \right)\] is the force which is the function of displacement x and \[dx\]is the small displacement, \[{x_i}\] and \[{x_f}\] is the initial and final position of the object.
Work-energy theorem, \[W = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2\]
where, m is the mass, \[{v_f}\] is the final velocity and \[{v_i}\] is the initial velocity.

Complete step by step answer:
As we can see the force acting on the particle is a variable force. We have the expression for the work done by the variable force,
\[W = \int\limits_{{x_i}}^{{x_f}} {F\left( x \right)\,dx} \]
Here, \[F\left( x \right)\] is the force which is the function of displacement x and \[dx\]is the small displacement, \[{x_i}\] and \[{x_f}\] is the initial and final position of the object.

Substituting \[F\left( x \right) = \left( {3x + 4} \right)\] in the above equation, we get,
\[W = \int\limits_0^4 {\left( {3x + 4} \right)\,dx} \]
\[ \Rightarrow W = \left( {\dfrac{{3{x^2}}}{2} + 4x} \right)_0^4\]
\[ \Rightarrow W = \left( {\dfrac{{3{{\left( 4 \right)}^2}}}{2} + 4\left( 4 \right)} \right) - \left( {0 + 0} \right)\]
\[ \Rightarrow W = 24 + 16\]
\[ \Rightarrow W = 40\,{\text{J}}\]

We have from the work-energy theorem; the work done by the force is equal to the change in kinetic energy of the object.
\[W = \Delta K\]
\[ \Rightarrow W = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2\]
Here, m is the mass, \[{v_f}\] is the final velocity and \[{v_i}\] is the initial velocity.
Substituting \[W = 40\,{\text{J}}\], \[m = 10\,{\text{kg}}\] and \[{v_i} = 2\,{\text{m}}{{\text{s}}^{ - 1}}\] in the above equation, we get,
\[40 = \dfrac{1}{2}\left( {10} \right)v_f^2 - \dfrac{1}{2}\left( {10} \right){\left( 2 \right)^2}\]
\[ \Rightarrow 40 + 20 = 5v_f^2\]
\[ \Rightarrow v_f^2 = 12\]
\[ \therefore {v_f} = 2\sqrt 3 \,{\text{m/s}}\]
Therefore, the final velocity of the object will be \[2\sqrt 3 \,{\text{m/s}}\].

So, the correct answer is option C.

Note: Students must note that the work done by the force is equal to the difference in the final kinetic energy and initial kinetic energy. But, since the change in kinetic energy is equal to a negative change in potential energy, the work done can also be expressed as difference in the initial potential energy and the final potential energy. If the body starts from the rest, the work done by the body is equal to the final kinetic energy.