To push a 25.0 kg crate up a frictionless incline, angled at $\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{°}$to the horizontal, a worker exerts a force of 209 N parallel to the incline. As the crate slides1.50 m, how much work is done on the crate by (a) the worker’s applied force, (b) the gravitational force on the crate, and (c) the normal force exerted by the incline on the crate? (d) What is the total work done on the crate?

It is given that,

Mass of the crate is m=25 kg

Applied force is $F_{app}=209\mathrm{N}$.

The angle between applied force and horizontal is$\theta =25°$

Horizontal displacement is$d=1.50\mathrm{m}$.

The problem deals with the work done which is the fundamental concept 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
• Concept of variable forces
• Work-force relation

Formulae:

The work done in general is given by

$W=F.dcos\varphi$ …(i)

Where,Fis force, dis displacement and $\varphi$is the angle between F and d.

The net work done is given by,

$W_{net}=W_n+W_{app}+W_g$ …(ii)

With the notation used in equation (i) the work done by the applied force is given by,

$$\begin{gathered} W_{app}=F_{app}.d\cos \varphi \\ =209\mathrm{N}\times 1.50\mathrm{m}\times \cos \left(0\right) \\ =313.5\mathrm{J} \\ \approx 314\mathrm{J} \end{gathered}$$

Hence, the work done by the applied force is 314J

$W_g=F_g.d\cos \varphi$

The angle between the x-component of the gravitational force and displacement d is

$\varphi =180°$

role="math" localid="1657180276736" $$\begin{gathered} W_g=mg\sin \theta \times d\times \cos \left(180\right) \\ =25\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \sin \left(25\right)\times 1.50\mathrm{m}\times \cos 180 \\ =25\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times (0.4226)\times 1.50\mathrm{m}\times (-1) \\ =-155.3122\mathrm{J} \\ \approx -155.0\mathrm{J} \end{gathered}$$

-

$W_N=N.d\cos \varphi$

role="math" localid="1657180592406" $$\begin{gathered} \mathrm{Normal}\mathrm{force}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{displacement}\mathrm{d}.\mathrm{Hence},\mathrm{\varphi }=90° \\ {W}_N=N.d\mathrm{cos\varphi } \\ =0 \end{gathered}$$

Hence, the work done by the normal force is 0 J

Using equation (ii) the work done is given by

$$\begin{gathered} W_{net}=-155+0+313.5 \\ =158.5\mathrm{J} \end{gathered}$$

Hence, the total work done is 158.5 J