Question

To pull a 50 kgcrate across a horizontal frictionless floor, a worker applies a force of 210 N, directed${\mathbf{20}}^{\mathbf{°}}$above the horizontal. As the crate moves 3.0 m, what work is done on the crate by (a) the worker’s force, (b) the gravitational force, and (c) the normal force? (d) What is the total work?

Short answer

1. Work done by the applied force is$W_{app}=590\mathrm{J}$
2. Work done by gravitational force is $W_g=0\mathrm{J}$.
3. Work done by the normal is $W_N=0\mathrm{J}$.
4. Total work done is $W_{net}=590\mathrm{J}$.

Step-by-step solution

Given information

It is given that,

Mass of the crate is m = 50 kg

Angle betweentheapplied force and horizontal is$\theta ={20}^{°}$

Horizontal displacement is x= 3.0 m

Determining the concept of power and work

The problem deals with the power and the work done. Work, energy, and power are the fundamental concepts of physics. Work is the displacement of an object when force is applied on it.

To solve this problem, use,

• x and y components of the force
• Work-energy theorem
• Work-force relation

Formula:

$W=F\times d$

where, F is force, d is displacement and W is the work done.

(a) Determining the work done by the applied force(Wapp)

Now,

$W_{app}=F\cos \theta \times d$

Using,$\theta ={20}^{°}$theapplied force is in the same direction of displacement, therefore,

$$\begin{gathered} W_{app}=210\mathrm{N}\times \cos 20\times 3\mathrm{m}\left(\frac{1\mathrm{J}}{1\mathrm{Nm}}\right) \\ =590\mathrm{J} \end{gathered}$$

Hence, the work done by the applied force is$W_{app}=590\mathrm{J}$

(b) Determining the work done by the gravitational force (Wg)

Gravitational force is perpendicular to the displacement x. So,$\theta ={90}^{°}$

$$\begin{gathered} W_g=F_g\cos \theta \times d \\ =F_g\cos 90\times d \\ =0\mathrm{J} \end{gathered}$$

Hence,the work done by gravitational force is $W_g=0\mathrm{J}$.

(c) Determining the work done by the normal (WN)

Normal force is perpendicular to the displacement x. Therefore, $\mathrm{\theta }={90}^{°}$.

$$\begin{gathered} W_N=N\cos \theta \\ =N\cos 90\times d \\ =0\mathrm{J} \end{gathered}$$

Hence, thework done by the normal is $W_N=0\mathrm{J}$.

(d) Determining the total work done (Wnet)

$$\begin{gathered} W_{net}=W_N+W_{app}+W_g \\ =0\mathrm{J}+590\mathrm{J}+0\mathrm{J} \\ =590\mathrm{J} \end{gathered}$$

Hence, the total work done is $W_{net}=590\mathrm{J}$.