Tag Archives: work on crate and work on the body

College Physics by Openstax Chapter 7 Problem 5

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 7.32.) 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 7.32 A man pushes a crate up a ramp.

Solution:

The Work Done by the Man on the Crate

The work WW that a force FF does on an object is the product of the magnitude FF of the force, times the magnitude dd of the displacement, times the cosine of the angle θ\theta between them. In symbols,

W=Fdcos⁡θW=Fd \cos \theta

In case where the work done by the man to the crate, the following values are given:

F= 500 Nd= 4 mθ= 0∘(Force is parallel to displacement)\begin{align*} F = & \ 500\ \text{N} \\ d = & \ 4\ \text{m} \\ \theta = & \ 0^{\circ} \color{Blue} \left( \text{Force is parallel to displacement} \right) \end{align*}

Substituting these values in the equation, we have

W= Fdcos⁡θW= (500 N)(4 m)cos⁡0∘W= 2000 N⋅m\begin{align*} W = & \ Fd \cos \theta \\ W = & \ \left( 500\ \text{N} \right) \left( 4\ \text{m} \right) \cos 0^{\circ} \\ W = & \ 2000\ \text{N} \cdot \text{m} \end{align*}

The work done by the man on his body

In this case, the force exerted is counteracted by the weight of the man. This force is directed upward. The displacement is still the 4.0 m along the inclined plane. The angle between the force and the displacement is 70 degrees.

W= Fdcos⁡θW= mgdcos⁡θW= (85.0 kg)(9.80 m/s2)(4.0 m)cos⁡70∘W= 1139.6111 N⋅m\begin{align*} W = & \ Fd \cos \theta \\ W = & \ mg d \cos \theta \\ W = & \ \left( 85.0\ \text{kg} \right) \left( 9.80\ \text{m/s}^2 \right)\left( 4.0\ \text{m} \right) \cos 70^{\circ} \\ W = & \ 1139.6111\ \text{N} \cdot \text{m} \end{align*}

The Total Work

The total work done by the man is the sum of the work he did on the crate and on his body.

WT=2000 N⋅m+1139.6111 N⋅mWT=3139.6111 N⋅mWT=3.14×103 N⋅mWT=3.14×103 J  (Answer)\begin{align*} W_{T} & = 2000\ \text{N}\cdot \text{m} + 1139.6111\ \text{N}\cdot \text{m} \\ W_{T} & = 3139.6111 \ \text{N}\cdot \text{m} \\ W_{T} & = 3.14 \times 10^{3} \ \text{N}\cdot \text{m} \\ W_{T} & = 3.14 \times 10^{3} \ \text{J}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}