Answer to Question #200217 in Mechanics | Relativity for Fhulu

Question

1.) A 1200 kg car is being driven up a 5° hill. The frictional force has a magnitude of 524 N. A force F is applied to the car by the road and propels the car forward. In addition to these forces, two other forces act on the car : its weight W and the normal force N directed perpendicular to the road surface. The length of the road up the hill is 290 m. Calculate the magnitude of F so that the net work done by all the forces acting on the car is 1150 kJ.

2.) A 1900 kg car experiences a combined force of air resistance and friction that has the same magnitude whether the car goes up or down a hill at 27 m/s. Going up a hill, the car's engine produces 47 hp more power to sustain the constant velocity than it does going down the hill. Calculate at what angle is the hill inclined above the horizontal.

Answer 1

1)

The forces N and mg cos($\theta$) act perpendicular to the displacement, so their work done is zero.

$\sum W=W_F+W_f+W_{mgsin\theta}=1150\ kJ$

$W_f=-fS=-524\cdot290=-151.96\ kJ$

$W_{mgsin\theta}=-1200\cdot9.8\cdot290sin5\degree=-297.34\ kJ$

$W_F=1150-W_f-W_{mgsin\theta}=1150+151.96+297.24=1599.2\ kJ$

$F=W_F/S=1599.2/290=5.5\ kN$

2)

F_R is the combine force of air resistance and friction, F_EU is the force provided by the engine when the car is going uphill, and F_ED is the force provided by the engine when the car is going downhill. Since the velocity is constant when the car is going up and down yield the following sum of force equations parallel to the incline.

Going uphill:

$\sum F_{||u}=F_{EU}-F_R-mgsin\theta=0$

$F_{EU}=F_R+mgsin\theta$

$\sum F_{||d}=F_{ED}-F_R+mgsin\theta=0$

$F_{ED}=F_R-mgsin\theta$

$\Delta P=P_U-P_D=47\ hp$

At constant speed we can write power as

$P=Fv$

$\Delta P=v(F_{EU}-F_{ED})=v(F_R+mgsin\theta-F_R+mgsin\theta)=2vmgsin\theta$

$sin\theta=\frac{\Delta P}{2mgv}=\frac{47\cdot745.7}{2\cdot2100\cdot9.8\cdot27}=0.00349$

$\theta=2\degree$