A body of mass $5 \mathrm{~kg}$ is moving with a momentum of $10 \mathrm{~kg} \mathrm{~ms}^{-1}$. Now a force of $2 \mathrm{~N}$ acts on the body in the direction of its motion for $5 \mathrm{~s}$. The increase in the Kinetic energy of the body is ___________ $\mathrm{J}$.

<p>The increase in kinetic energy can be found using the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.</p> <p>The work done by a force is given by the equation:</p> <p>$ W = F \cdot d $</p> <p>where ( F ) is the force and ( d ) is the distance over which the force is applied. </p> <p>However, we don&#39;t have the distance in this problem. But we do know that the force is applied for a time of 5 seconds, and that the initial momentum of the body is 10 kg m/s. We can use these facts to find the work done.</p> <p>First, we can use the equation for force, ( F = ma ), to find the acceleration of the body:</p> <p>$a = \frac{F}{m} = \frac{2 \, \text{N}}{5 \, \text{kg}} = 0.4 \, \text{m/s}^2 $</p> <p>Then, we can use the equation for distance in uniformly accelerated motion, ( $d = v_i t + \frac{1}{2} a t^2 $), where ( $v_i$ ) is the initial velocity of the body. We can find ( $v_i $) from the initial momentum and the mass of the body:</p> <p>$ v_i = \frac{p}{m} = \frac{10 \, \text{kg m/s}}{5 \, \text{kg}} = 2 \, \text{m/s} $</p> <p>Substituting ( $v_i$ ), ( a ), and ( t ) into the equation for ( d ) gives:</p> <p>$ d = 2 \, \text{m/s} \cdot 5 \, \text{s} + \frac{1}{2} \cdot 0.4 \, \text{m/s}^2 \cdot (5 \, \text{s})^2 = 10 \, \text{m} + 5 \, \text{m} = 15 \, \text{m} $</p> <p>Finally, we can substitute ( F ) and ( d ) into the equation for work to find the increase in kinetic energy:</p> <p>$ \Delta KE = W = F \cdot d = 2 \, \text{N} \cdot 15 \, \text{m} = 30 \, \text{J} $</p> <p>So, the increase in the kinetic energy of the body is 30 J.</p>