A 1460-kg car is being driven up a 10.0 ° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of 497 N. A force F is applied to the car by the road and propels the car forward. In addition to these two forces, two other forces act on the car: its weight W and the normal force FN directed perpendicular to the road surface. The length of the road up the hill is 209 m. What should be the magnitude of F, so that the net work done by all the forces acting on the car is 149 kJ?
Fc=mg=1460kg * 9.8N/kg=14308N @ 10deg.
Fp = 14308*sin10 = 2484.56N = Force parallel to the plane downward.
Fv = 14308*cos10 = Ff = .63N = Force perpendicular to the plane and acting downward.
Ff = 497N = Force of friction acting
opposite to the motion.
W = Fn*d = 149000J.
Fn * 209 = 149000,
Fn = 149000 / 209 = 712.92 = Net force.
Fn = Fap - Fp - Ff = 712.92,
Fap - 2484.56 - 497 = 712.92,
Fap - 2981.56 = 712.92,
Fap = 712.92 + 2981.56 = 3695N = Force
applied.
CORRECTION:
Fv = 14308*cos10 = 14090.63N = Force perpendicular to the plane and acting
downward.
To find the magnitude of force F, we need to calculate the net work done by all the forces acting on the car:
Net work (W_net) = Work done by F + Work done by Friction + Work done by Weight
The work done by any force is given by the equation:
Work = Force * Distance * Cosine(theta)
where theta is the angle between the force and the direction of motion.
Let's calculate each of the three terms:
1. Work done by F:
The force F is applied parallel to the direction of motion, so theta = 0°.
Work_F = F * Distance * Cosine(0°) = F * Distance
2. Work done by Friction:
The frictional force has a magnitude of 497 N and acts opposite to the motion, so theta = 180°.
Work_Friction = Friction * Distance * Cosine(180°) = -Friction * Distance
Note: Since the force and distance have opposite directions, the work done by friction is negative.
3. Work done by Weight:
The weight W acts vertically downward, perpendicular to the direction of motion, so theta = 90°.
Work_Weight = Weight * Distance * Cosine(90°) = 0
Note: The work done by weight is zero because the weight is perpendicular to the direction of motion.
Now, let's substitute the values given in the problem:
W_net = Work_F + Work_Friction + Work_Weight
W_net = F * Distance - Friction * Distance + 0
W_net = (F - Friction) * Distance
Given that W_net = 149 kJ and Distance = 209 m:
149 kJ = (F - Friction) * 209 m
Now, let's substitute the value of Friction (497 N) and solve for F:
149 kJ = (F - 497 N) * 209 m
To convert kilojoules to joules, multiply by 1000:
149,000 J = (F - 497 N) * 209 m
Divide both sides by 209 m:
F - 497 N = 149,000 J / 209 m
F - 497 N = 713.87 N
Add 497 N to both sides:
F = 713.87 N + 497 N
F ≈ 1210 N
Therefore, the magnitude of force F should be approximately 1210 N so that the net work done by all the forces acting on the car is 149 kJ.