Step 1

We are given the force exerted by a motor on the cable as $F = \left( {600 + 2{s^2}} \right)\;{\rm{N}}$, the mass of a crate as $m = 100\;{\rm{kg}}$ and the distance covered by a crate from the ground as $s = 15\;{\rm{m}}$.

We are asked to determine the speed of a crate.

 
Step 2

To calculate the speed of a crate $\left( {{v_2}} \right)$, we need to apply the principle of work and energy as:

\[\begin{array}{c} {K_1} + \sum {U_{1 - 2}} = {K_2}\\ \frac{1}{2}mv_1^2 + \sum {U_{1 - 2}} = \frac{1}{2}mv_2^2 \end{array}\]   ...... (1)

Here, ${v_1}$ is the initial velocity, ${v_2}$ is the final velocity and $\sum {U_{1 - 2}}$ is the work done by all the forces.

 
Step 3

As the crate is initially at rest condition so the value of ${v_1}$ will be zero $\left( {{v_1} = 0} \right)$.

Since, the motor exerts the force on the cable, the forces acting in the cable at point C are two times the motor exerts the force on the cable. Hence, the work done by the force exerted on motor can be calculated as:

\[\begin{array}{c} {\left( {{U_{1 - 2}}} \right)_1} = 2\int\limits_0^{15} {Fds} \\ {\left( {{U_{1 - 2}}} \right)_1} = 2\int\limits_0^{15} {\left( {600 + 2{s^2}} \right)ds} \\ {\left( {{U_{1 - 2}}} \right)_1} = 2\left( {600s + \frac{{2{s^3}}}{3}} \right)_0^{15} \end{array}\]

Substitute the limits in s, we get:

\[\begin{array}{c} {\left( {{U_{1 - 2}}} \right)_1} = 2\left[ {\left( {600 \times \left( {15} \right) + \frac{{2 \times {{\left( {15} \right)}^3}}}{3}} \right) - \left( {600 \times \left( 0 \right) + \frac{{2 \times {{\left( 0 \right)}^3}}}{3}} \right)} \right]\\ {\left( {{U_{1 - 2}}} \right)_1} = 22500\;{\rm{J}} \end{array}\]
 
Step 4

Now, the work done by the crate can be calculated as:

\[\begin{array}{c} {\left( {{U_{1 - 2}}} \right)_2} = - mgs\\ {\left( {{U_{1 - 2}}} \right)_2} = - \left( {100\;{\rm{kg}}} \right) \times \left( {9.81\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right. } {{{\rm{s}}^2}}}} \right) \times \left( {15\;{\rm{m}}} \right)\\ {\left( {{U_{1 - 2}}} \right)_2} = - 14715\;{\rm{J}} \end{array}\]

Here, a negative sign indicates that the weight of a crate acts downward.

Hence, the work done by all the forces will be:

\[\begin{array}{c} \sum {U_{1 - 2}} = {\left( {{U_{1 - 2}}} \right)_1} + {\left( {{U_{1 - 2}}} \right)_2}\\ \sum {U_{1 - 2}} = \left( {22500\;{\rm{J}}} \right) + \left( { - 14715\;{\rm{J}}} \right)\\ \sum {U_{1 - 2}} = 7785\;{\rm{J}} \end{array}\]

Substitute all the values in equation (1), we get:

\[\begin{array}{c} \frac{1}{2}mv_1^2 + \sum {U_{1 - 2}} = \frac{1}{2}mv_2^2\\ \frac{1}{2} \times \left( {100\;{\rm{kg}}} \right) \times {\left( 0 \right)^2} + \left( {7785\;{\rm{J}}} \right) = \frac{1}{2} \times \left( {100\;{\rm{kg}}} \right) \times v_2^2\\ \left( {7785\;{\rm{J}}} \right) = \left( {50\;{\rm{kg}}} \right) \times v_2^2\\ {v_2} = 12.47\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. } {\rm{s}}} \end{array}\]