SOLUTIONS TO THE SECOND HOUR EXAM
1. Newtons's first law states that an object at rest stays at rest unless acted on by outside forces; an object in motion stays in constant motion unless acted on by
The second law statest that F = ma.
The third law states that for every action force, there is an equal an opposited reaction force
a) W = mg, so on Earth this object weighs 100kg x 10 m/s/s = 1000N.
b) On a planet where g = 25 m/s/s, W =mg = 100kg x 25 m/s/s = 2500 N.
3. She gains more speed in the first second of fall. When she starts falling, her speed is near zero, so the amount of opposing air friction is small. (Remember that air
friction is a force that increases as the speed of fall through the air increases). So in the first second, the dominant force on her is gravity acting down, and she
experiences little counter force, so her acceleration down is greatest early in the trip.
Near the end of the trip, she has reached nearly her terminal velocity, i.e., her maximum velocity of fall, and thus covers more distance in the last one second interval
than in the first one second interval, when she is starting from rest.
a) The key phrase in this question is that the crate is moving at constant speed, that we know it is not accelerating. If the crate is not accelerating, then there are no
net forces acting on it, so the force of friction opposing the crate must equal the pulling force. Thus, the force of friction is 100 N (this is just like the friction
experiments we did in class).
b) If the crate has a mass of 30 kg, it has a weight of 300N, and this is the force pushing the crate and surface together. Thus, the coefficient of friction is found
frictional force = (coefficient of friction) x (normal force)
coefficient of friction = (frictional force)/(normal force) = 100 N/ 300 N = 1/3
5. The common misconception is that the rocket fuel pushes against something to move the rocket. You can think of this is simply a matter of conservation of momentum,
where the fuel leaves the rocket at high speed, and in order to conserve momemntum the rocket moves forward at high speed. Or, you can argue that the rocket exerts a force
on the fuel to force it out of the back of the rocket, and the fuel then exerts and equal and forward directed force pushing the rocket through space.
6. Impulse is the force exerted on an object multiplied by the time in which the force is in contact with the object; this is equivalent to the change in momentum
experienced by the object. Momentum is the vector product of mass x velocity. Conservation of momentum states that in the absence of outside forces, the momentum if a
system does not change.
7. Each bullet has the same momentum, since we specify they have the same mass and velocity. In order to cause the rubber bullet to bounce backward, the can must exert a
greater force on the bullet, since this involves a much greater force than with regard to the steel bullet.
a) The potential energy is mgh. The kinetic energy is zero.
b) Just before impact, the kinetic energy is 1/2 mv2, the potential energy is zero.
c) Since the total energy never changes, we can equate total energy at the top to total energy at the bottom. This yields:
KE at top + PE at top = KE at bottom + PE at bottom
mgh = 1/2mv2
v = sqrt (2gh) = sqrt (2 x 10 m/s/s x 5 m) = 10 m/s
a) W = F x d = 100 kg x 10 m/s/s/ x 30 m = 30,000 J.
b) If I have to push the mass up a ramp three times as long as the height of the building, I need to use 1/3 as much force. In the first instance I have to use mg or 1000N
of force, here I have to use 1/3 that amount or 333 N of force (but I have to push it through three times the distance).
Alternately, you can answer by saying that you have to do the same amount of work, i.e., 30,000J. Recalling that Work = force x distance, this means that you have to exert
a force equal to:
F = Work/distance = 30,000J/90 m = 333 N
David B. Slavsky
Loyola University Chicago
Cudahy Science Hall, Rm. 404
1032 W. Sheridan Rd.,
Chicago, IL 60660
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