Why Does a Car Traveling at Constant Velocity Still Do Work?

[Cpt. Bob]

Im studying on my own and don't have the benefit of input from an instructor, so any help with these few questions would be very appreciated.

1. If a car traveling at a constant velocity has a net force of 0, why can it be said that work is being done on the car when Work equals force multiplied by distance?

2. A golf ball is thrown at and bounces off a very massive bowling ball that is initially at rest. Which ball, after impact has greater momentum, and which has greater kinetic energy?

On question two, I think the answer is the bowling ball has greater momentum, since the golf ball changed velocity double what it would have relative to the bowling ball if it had not bounced, and since momentum is conserved, twice the momentum would be imparted into the bowling ball in order for net momentum = 0. And since KE=.5mv^2, I would think the golf ball would have more kinetic energy, having a much higher velocity. Am I correct here, or am I missing something. Thanks.

 

[quantumdude]

Originally posted by Cpt. Bob
1. If a car traveling at a constant velocity has a net force of 0, why can it be said that work is being done on the car when Work equals force multiplied by distance?

In the real world, there is air resistance to the motion of a car. That is a force against the motion, and so an additional force has to be supplied by the engine to make up for it. separately, the two forces do work on the car. The work done by the air is negative, and the work done by the engine is positive. For zero net force, the work done by each force has precisely the same magnitude, so they add up to zero.

2. A golf ball is thrown at and bounces off a very massive bowling ball that is initially at rest. Which ball, after impact has greater momentum, and which has greater kinetic energy?

On question two, I think the answer is the bowling ball has greater momentum, since the golf ball changed velocity double what it would have relative to the bowling ball if it had not bounced, and since momentum is conserved, twice the momentum would be imparted into the bowling ball in order for net momentum = 0. And since KE=.5mv^2, I would think the golf ball would have more kinetic energy, having a much higher velocity. Am I correct here, or am I missing something. Thanks.

I would do this explicitly.

Let:
m=mass of golf ball
M= mass of bowling ball
v_i=initial velocity of golf ball
v_f=final velocity of golf ball
V=final velocity of bowling ball

Conservation of Momentum:
mv_i=mv_f+MV

If the collision is elastic, then we also have...

Conservation of Kinetic Energy:
(1/2)mv_i²=(1/2)mv_f²+(1/2)MV²

You have two equations and two unknowns (v_f and V).

 

[M98Ranger]

1. When a car is traveling at a constant velocity, there is no change in its speed or direction. This means that the net force acting on the car is equal to 0, since there is no acceleration. However, even though the net force is 0, work can still be done on the car. This is because work is defined as the product of force and distance, not just force alone. So, even though there is no net force, the car is still moving and covering a certain distance, which means work is being done on it.

2. Your understanding of the situation is correct. The bowling ball, being more massive, will have a greater momentum after the impact. This is because momentum is directly proportional to mass. And as you correctly stated, since momentum is conserved in a collision, the golf ball will transfer twice its momentum to the bowling ball. However, when it comes to kinetic energy, the golf ball will have a greater amount. This is because kinetic energy is directly proportional to the square of velocity, and the golf ball will have a higher velocity after the collision due to the elastic nature of the collision. So, the golf ball will have a greater kinetic energy compared to the bowling ball.