Summary







































klm
You push on a car and it doesn't move. Why doesn't it?
Person strains to push a car
A. Because objects at rest tend to stay at rest.
B. Because the car pushes back on you with equal force.
C. Because the net force on the car is zero.
D. Because your pushing force is insufficient to overcome the car's inertia.
Answer



















PoP5 5.03
If 75.0 N is required to get a 25.0-kg block moving horizontally, and 60.0 N is required to maintain constant speed, what are µs and µk?
A. 0.417, 0.300
B. 0.188, 0.150
C. 0.306, 0.245
D. 0.525, 0.360
Answer



















klm
What applied force is needed to accelerate 50.0 kg at 2.00 m/s2 if the frictional force is 150 N?

A. 100 N
B. 250 N
C. −50 N
D. 50 N
Answer



















klm
A 2.00-kg book sliding along a horizontal table has a constant frictional force of 4.00 N acting on it. If it has an initial speed of 8.00 m/s, for how much time will it slide before stopping?
A. 1.00 s
B. 2.00 s
C. 3.00 s
D. 4.00 s
Answer



















pop4 5.q.6
A book is given a brief push to make it slide up a rough incline. Compare the time intervals for it to go up (Δtup) and come down (Δtdown).
A. Δtup > Δtdown
B. Δtup = Δtdown
C. Δtup < Δtdown
D. It depends upon relative values of µs and µk.
Answer



















PSE10e Ex.6.4
Determine the angle θ at which a curved roadway should be banked such that no friction is needed to round the curve without skidding.
car on banked road
A. sin−1 (m g/r)
B. sin−1 (g r)
C. tan−1 (v²/r g)
D. tan−1 (r m g)

Answer



















PSE9 6.16b
A 500-kg roller coaster car goes over a hill of radius 15.0 m. What is the maximum speed the car can have at point B and still remain on the track?
roller coaster with circular hill
A. 24.7 m/s
B. 32.2 m/s
C. 7.32 m/s
D. 12.1 m/s
Answer



















PSE6 6.18
A 0.400-kg ball is swung in a vertical circular path on a string 0.500 m long. If its speed is v = 4.00 m/s at the top of the circle, what is the tension in the string there?
A. 8.88 N
B. 12.8 N
C. 3.92 N
D. 16.7 N
Answer



















PSE6 6.26
At what Earth rotation period would an object at the equator have zero apparent weight? RE = 6.37×106 m.
gravitational and centrifugal force vectors for Earth A. 22.6 h
B. 1.41 h
C. 12.2 h
D. 0.117 h
Answer



















klm
An object is subject to a resistive force that is proportional to velocity, R = −bv. If b = 0.00400 N/m/s and m = 1.45 g, what is v at t = 0.200 s?
A. 1.51 m/s
B. 2.00 m/s
C. 2.76 m/s
D. 3.55 m/s
Answer



















PoP5 5.34
A 9.00-kg object falls from rest through a viscous medium and experiences a resistive force R = −bv. It reaches one-half terminal speed in 5.54 s. How far has the object traveled in 5.54 s?
A. 49.9 m
B. 88.5 m
C. 121 m
D. 276 m
Answer



















klm
Two objects are subject to a resistive force that is proportional to velocity, D = −½CρAv². Object 1 has a C1 = 0.800 and object 2 has a C2 = 0.400. If each has the same mass, which object has the higher terminal velocity?
A. Object 1
B. Object 2
C. The terminal velocities are the same because the masses are the same.
D. The terminal velocities are the same because D is proportional to v² for each.
Answer



















pse6 6.32
A 80.0-kg sky diver jumps from a slow-moving airplane and reaches a terminal speed of 50.0 m/s. What is the acceleration of the sky diver when her speed is 30.0 m/s?
A. 3.53 m/s2
B. 6.27 m/s2
C. 5.88 m/s2
D. 9.80 m/s2
Answer



















answer

Person strains to push a car
C. Because the net force on the car is zero.

The static friction force between the tires and the ground cancels the force of your push, making the net force on the car equal to zero.



















answer

C. 0.306, 0.245
solution equation



















answer

B. 250 N
solution equation



















answer

D. 4 s
solution equation



















answer

C. Δtup < Δtdown

Note that if there were no friction, the two time intervals (up the ramp and then down the ramp) would be identical.



















answer
car on banked road
C. tan−1 (v²/r g)

solution equation



















answer

D. 12.1 m/s
Picture (908x426, 9.2Kb)



















answer

A. 8.88 N
Picture (973x245, 6.6Kb)



















answer

B. 1.41 h
solution equation



















answer

A. 1.51 m/s
solution equation



















answer

C. 121 m
solution equation



















answer

B. Object 2
One way to look at it is to say that a small C corresponds to a small resistive force, thus a high terminal speed. Another strategy is to use a ratio:
solution equation



















answer

B. 6.27 m/s2
solution equation



















 
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