A 45 kgblock of ice slides down a frictionless incline 1.5mlong and 0.91 mhigh. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker’s force. How much work is done on the block by (b) the worker’s force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?

It is given that,

Mass of the ice block is$m=45\mathrm{kg}$

Length of incline is $L=1.5m$

Height of the incline is h =0.91m.

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 to it. Also,it involves Newton’s second law of motion which can be used to find net force. First, resolve the forces acting on the block.Find the angle of incline from the given length and height of the incline. Then plugging the values for each force and displacement, find the work done.

Formulae:

$$\begin{gathered} W=Fd\cos \left(\varphi \right) \\ \theta ={\sin }^{-1}\left(\frac{h}{L}\right) \\ F=ma \end{gathered}$$

From the above figure,calculate the work done by each force,

Firstly, find the angle from the given lengths of the incline,

$$\begin{gathered} \sin \theta =\frac{0.91}{1.5} \\ =0.6066 \\ \theta ={\sin }^{-1}(0.6066) \\ =37.34° \end{gathered}$$

Apply Newton’s second law to find the applied force,

$$\begin{gathered} F_{net}=ma \\ =F_a-mg\sin \theta \\ ma=F_a-mg\sin \theta \end{gathered}$$

As the block moves with constant velocity, the acceleration is zero,

$$\begin{gathered} 0=F_a-mg\sin \theta \\ {F}_a=mg\sin \theta \\ =45\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \sin (37.34) \\ =267.54\mathrm{N} \\ =2.7\times {10}^2\mathrm{N} \end{gathered}$$

Hence, magnitude of the worker’s force is$F_a=2.7\times {10}^2\mathrm{N}$

Force by the worker is opposite to the displacement, so work done as,

$W_1=F_{a}L\cos \left(\theta \right)$

$\varphi$is the angle between force and the displacement,

$$\begin{gathered} W_1=-2.7\times {10}^2\mathrm{N}\times 1.5\mathrm{m}\times \cos \left(180\right) \\ =-405 \\ =-4.0\times {10}^2\mathrm{J} \end{gathered}$$

Hence, work done on the block by the worker’s force is ${\mathrm{W}}_1=-4.0\times {10}^2\mathrm{J}$.

Here, gravitational force and the displacement is in the same direction downward.Hence,

$$\begin{gathered} W_2=\mathrm{mg}\left(\mathrm{h}\right)\cos \left(0\right) \\ =45\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 0.91\mathrm{m}\times \cos \left(0\right) \\ =401.31\mathrm{J} \\ =4.0\times {10}^2\mathrm{J} \end{gathered}$$

Hence, work done on the block by the gravitational force is $4.0\times {10}^{2}J$.

Here,the normal force and displacement are perpendicular to each other,

$$\begin{gathered} {\mathrm{W}}_3=\mathrm{mg}\cos \left(\mathrm{\theta }\right)\mathrm{Lcos}\left(\mathrm{f}\right) \\ =45\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times \cos (37.34)\times 1.5\mathrm{m}\times \left(90\right) \\ =0\mathrm{J} \end{gathered}$$

Hence, work done on the block by the normal force of the incline is 0 J.

As the block is moving with a constant velocity, there is no acceleration.Hence, net force is 0.

Therefore, work done by the net force is 0J.