91Ó°ÊÓ

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$\mathrm{θ}=25°$

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

$W=F.dcos\mathrm{Ï•}$ …(¾±)

Where,Fis force, dis displacement and $\mathrm{Ï•}$is the angle between F and d.

The net work done is given by,

$W_{net}=W_n+W_{app}+W_g$ …(¾±¾±)

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

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

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

$W_g=F_g.d\cos \mathrm{Ï•}$

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

$\mathrm{ϕ}=180°$

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

-

$W_N=N.d\cos \mathrm{Ï•}$

role="math" localid="1657180592406" $$\begin{gathered} \mathrm{Normal}\mathrm{force}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{displacement}\mathrm{d}.\mathrm{Hence},\mathrm{Ï•}=90° \\ {W}_N=N.d\mathrm{³¦´Ç²õÏ•} \\ =0 \end{gathered}$$

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