91Ó°ÊÓ

The power output of the force pulling an object at a steady speed is given by dot product of force vector and velocity vector is

$$\begin{gathered} P=\stackrel{→}{F}.\stackrel{→}{v} \\ P=Fv\cos \left(\mathrm{Î\pm }\right) \end{gathered}$$

Draw the free body diagram

Here, the pulling force applied by the driver is F in the direction upward along the incline. Normal force inclined on bail is N acting perpendicular to incline. Gravitational force on bail is mg.

Coefficient of friction is f, and friction coefficient is${\mathrm{μ}}_K$.

Angle of inclination is$\mathrm{θ}$and bale is moving at a constant velocity v.

Component of force of gravity along the incline is $mg\sin \left(\mathrm{θ}\right)$and Component of force of gravity normal the incline is$mg\cos \left(\mathrm{θ}\right)$

Newton's second law of motion if object is moving at a constant velocity net force must be zero.

$F_{net}=0$

Net force normal to incline is zero.

$N=mg\cos \left(\mathrm{θ}\right)$

Here, net force along the incline is zero.

$F=f=mg\sin \left(\mathrm{θ}\right)=0$

Rearrange the above equation

role="math" localid="1647802991645" $F=f+mg\sin \left(\mathrm{θ}\right)..........\left(1\right)$

Friction force is f given as

$f={\mathrm{μ}}_{K}N................\left(2\right)$

Substitute $mg\cos \left(\mathrm{θ}\right)$for N in equation $\left(2\right)$

$f={\mathrm{μ}}_{K}mg\cos \left(\mathrm{θ}\right)$

Substitute this value in equation $\left(1\right)$

$F={\mathrm{μ}}_{K}mg\cos \left(\mathrm{θ}\right)+mg\sin \left(\mathrm{θ}\right)..................\left(3\right)$

Power output of tractor is

$$\begin{gathered} P=Fv\cos \left(\mathrm{Î\pm }\right) \\ P=mg({\mathrm{μ}}_K\cos \left(\mathrm{θ}\right)+\sin \left(\mathrm{θ}\right)v\cos \left(\mathrm{Î\pm }\right)..............(4) \end{gathered}$$

Force and velocity are in same direction so $\mathrm{Î\pm }=0°$

$$\begin{gathered} P=\left(150kg\right)(9.8m/s^2)(0.45\cos \left(15°\right)+\sin \left(15°\right)(5.0km/hr\left)\right(\cos \left(0°\right) \\ P=1415.87W\left(\frac{1KW}{{10}^{3}W}\right) \\ P=1.4kW \end{gathered}$$