Step 1

We are given the mass of the crate is $m = 100\;{\rm{kg}}$, the final speed of the crate is $v = 8\;{\rm{m/s}}$, and the coefficient of kinetic friction between the crate and surface is ${\mu _k} = 0.2$.

We are asked to determine the distance of crate slides in order to attain a speed of $v = 8\;{\rm{m/s}}$.

 
Step 2

The diagram for the system is shown as:

We have the force 1 on crate is ${F_1} = 400\;{\rm{N}}$.

We have the angle of force 1 from horizontal is ${\theta _1} = 30^\circ $.

We have the force 2 on crate is ${F_2} = 500\;{\rm{N}}$.

We have the angle of force 2 from horizontal is ${\theta _1} = 45^\circ $.

We have the initial velocity of the crate is $u = 0$.

 
Step 3

The formula to calculate the normal reaction on the crate by equilibrium condition is,

\[N + {F_2}\sin {\theta _2} - mg - {F_1}\sin {\theta _1} = 0\]

Here, g is the gravitational acceleration having a standard value of $9.81\;{\rm{m/}}{{\rm{s}}^2}$.

 
Step 4

Substitute the values in the above expression.

\[\begin{array}{c} \left[ \begin{array}{l} N + \left( {500\sin 45^\circ } \right)\;{\rm{N}} - \left( {100\;{\rm{kg}}} \right)\left( {9.81\;{\rm{m/}}{{\rm{s}}^2}} \right) \times \left( {\frac{{1\;{\rm{N}}}}{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)\\ - \left( {400\sin 30^\circ } \right)\;{\rm{N}} \end{array} \right] = 0\\ N = 827.45\;{\rm{N}} \end{array}\]
 
Step 5

The formula to calculate the distance of crate slides in order to attain a speed of $v = 8\;{\rm{m/s}}$ by energy conservation is,

\[\begin{array}{c} {T_1} + {F_1}\cos {\theta _1}s + {F_2}\cos {\theta _2}s - \left( {{\mu _k}N} \right)s = {T_2}\\ \frac{1}{2}m{u^2} + {F_1}\cos {\theta _1}s + {F_2}\cos {\theta _2}s - \left( {{\mu _k}N} \right)s = \frac{1}{2}m{v^2} \end{array}\]
 
Step 6

Substitute the values and integrate the above expression.

\[\begin{array}{c} \left[ \begin{array}{l} \frac{1}{2}\left( {100\;{\rm{kg}}} \right){\left( 0 \right)^2} + \left[ {\left( {400\cos 30^\circ } \right)\;{\rm{N}} \times \left( {\frac{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1\;{\rm{N}}}}} \right)} \right]\left( s \right)\\ + \left[ {\left( {500\cos 45^\circ } \right)\;{\rm{N}} \times \left( {\frac{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1\;{\rm{N}}}}} \right)} \right]\left( s \right)\\ - \left( {0.2} \right)\left( {827.45\;{\rm{N}}} \right)\left( {\frac{{1\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}{{1\;{\rm{N}}}}} \right) \times \left( s \right) \end{array} \right] = \frac{1}{2}\left( {100\;{\rm{kg}}} \right){\left( {8\;{\rm{m/s}}} \right)^2}\\ \left[ {0 + 346.41\left( s \right) + 353.55\left( s \right) - 165.49\left( s \right)} \right]{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^2} = \frac{1}{2}\left( {100\;{\rm{kg}}} \right){\left( {8\;{\rm{m/s}}} \right)^2}\\ s = 5.98\;{\rm{m}} \end{array}\]