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?

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

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.

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$W_{app}=590\mathrm{J}$

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 $W_g=0\mathrm{J}$.

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 $W_N=0\mathrm{J}$.

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

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