Question

Calculate the work done by an 85.0-kg man who pushes a crate 4.00 m up along a ramp that makes an angle of 20.0º with the horizontal. (See Figure 7.35.) He exerts a force of 500 N on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.

Figure 7.35 A man pushes a crate up a ramp

Short answer

Work done by a man to push a crate \(4{\rm{ m}}\) up on the ramp is \(3140.71{\rm{ J}}\).

Step-by-step solution

Definition of Concept

Work: Work is the product of the component of force in the displacement direction and the magnitude of the displacement.

Calculate the work done

Considering the given information:

Mass\(85.0{\rm{ kg}}\).

Cart’s displacement\(d = 4.00{\rm{ m}}\).

Angle of inclination of the slope \(\theta = 20.0^\circ \).

Force exerted on the cart\(F = 500{\rm{ N}}\).

The height the cart gets up on the ramp is,

\(\begin{aligned}{}h &= d\sin \theta \\ &= \left( {4.00{\rm{ m}}} \right) \times \sin \left( {20^\circ } \right)\\ &= 1.368{\rm{ m}}\end{aligned}\)

The work done by the man is,

\(\begin{aligned}{}W &= \left( \begin{aligned}{l}{\rm{increase in potential }}\\{\rm{energy of the cart}}\end{aligned} \right) + \left( \begin{aligned}{}{\rm{work done in displacing }}\\{\rm{the cart}}\end{aligned} \right)\\ &= mgh + Fd\end{aligned}\)

Here,\(g\)is the acceleration due to gravity\(\left( {g = 9.81{\rm{ m}}/{{\rm{s}}^2}} \right)\).

Putting all known values,

\(\begin{aligned}{}W &= \left( {85.0{\rm{ kg}}} \right) \times \left( {9.81{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {1.368{\rm{ m}}} \right) + \left( {500{\rm{ N}}} \right) \times \left( {4.00{\rm{ m}}} \right)\\& = 3140.7{\rm{ J}}\end{aligned}\)

Therefore, work done by a man to push a crate \(4{\rm{ m}}\) up on the ramp is \(3140.71{\rm{ J}}\).