SciencePhysicsMastering Physics 5
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Mastering Physics 5

Cayden Brooks avatar Cayden Brooks
25 July 2022
4.7 (114 reviews)
19 test answers

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question
Positive x axis to the right, positive y axis upward
Positive x axis to the right, positive y axis upward
answer
The box is the important object in this problem. Choose a coordinate system that will simplify your work, and then draw a free-body diagram for the box. You should generally align the positive x axis with the direction in which your object of interest is moving, or the direction in which it would move if there were no friction. If the figure (Figure 1) depicts the situation, which of the following describes the coordinate system that you should use? Positive x axis to the right, positive y axis downward Positive x axis to the left, positive y axis upward Positive x axis to the left, positive y axis downward Positive x axis to the right, positive y axis upward
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f = 208 N
f = 208 N
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What is the magnitude f of the force of friction?
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n = 414 N
n = 414 N
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Find the magnitude n of the normal force.
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F1<F2?2F1
F1
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The baggage handler now pushes a second box, identical to the first, so that it accelerates at a rate of 2a. How does the magnitude of the force F2 that the handler applies to this box compare to the magnitude of the force F1 applied to the first box? 0?F2<1/2F1 1/2F1?F22F1
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1/2F1?F3<F1
1/2F1?F3
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Now assume that the baggage handler pushes a third box (Figure 2) of mass m/2 so that it accelerates at a rate of 2a. How does the magnitude of the force F3 that the handler applies to this box compare to the magnitude of the force F1 applied to the first box? 0?F3<1/2F1 1/2F1?F32F1
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increases.
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As the ramp angle increases, the force of static friction remains the same. increases. decreases.
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zero
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With the crate stationary on a horizontal ramp, the force of static friction is directed to the left. zero. directed to the right.
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? = 35?
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What is the maximum ramp angle that still allows the crate to remain at rest? (Make sure the coefficient of friction is 0.7.) SIMULATION Q NOT SURE IF NEEDED
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The maximum angle decreases.
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In the previous part, you determined the maximum angle that still allows the crate to remain at rest. If the coefficient of friction is less than 0.7, what happens to this angle? (Note that you can adjust the coefficient of friction by clicking on the More Features tab near the top of the window and then using the slider bar in the right panel.) The maximum angle increases. The maximum angle decreases. The maximum angle remains the same. SIMULATION Q NOT SURE IF NEEDED
question
The maximum angle does not depend on the mass.
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How does the maximum angle for which the crate can remain at rest on the ramp depend on the mass of the crate? The maximum angle increases as the mass increases. The maximum angle does not depend on the mass. The maximum angle decreases as the mass increases.
question
The force of static friction when there is no applied force is greater than the case when there is an applied force.
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How do the two forces of static friction compare? The force of static friction when there is no applied force is equal to the case when there is an applied force. The force of static friction when there is no applied force is greater than the case when there is an applied force. The force of static friction when there is no applied force is less than the case when there is an applied force.
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Yes, but only for a specific applied force directed up the ramp.
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For a stationary crate (with a coefficient of friction of 0.7) on the 15? ramp, can the force of static friction ever be zero? Yes, but only for a specific applied force directed down the ramp. Yes, but only for a specific applied force directed up the ramp. No
question
The minimum push needed to get the crate to slide up the ramp is greater than that to get the crate to slide down the ramp.
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Slowly adjust the applied force (pushing both up and down the ramp) until the crate begins to move. Determine the minimum strength of the pushing force needed to accelerate the crate up the ramp and the minimum strength of the pushing force needed to accelerate the crate down the ramp. How do these two minimum strengths compare to each other? The minimum push needed to get the crate to slide up the ramp is less than that to get the crate to slide down the ramp. The minimum push needed to get the crate to slide up the ramp is the same as that to get the crate to slide down the ramp. The minimum push needed to get the crate to slide up the ramp is greater than that to get the crate to slide down the ramp.
question
A=D>H=E>F=B=G> C
A=D>H=E>F=B=G> C
answer
Rank the ropes on the basis of the force each exerts on the crate immediately to its left.
question
The frictional force points up the incline.
The frictional force points up the incline.
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The crate is at rest on the incline. What can you say about the force of friction acting on the crate? The frictional force points up the incline. The frictional force points down the incline. The frictional force is zero.
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The frictional force points down the incline.
The frictional force points down the incline.
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A physicist attempts to push the crate up the incline. The physicist senses that if he applies slightly more force the crate will move up the incline but cannot muster enough strength to get the motion started. What can you say now about the force of friction acting on the crate? The frictional force points up the incline. The frictional force points down the incline. The frictional force is zero.
question
is greater than
is greater than
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The first physicist gets a second physicist to help. They both push on the crate, parallel to the surface of the incline, and it moves at constant speed up the incline. How does the force exerted by the two physicists on the crate compare with the force of friction on the crate? F two physicists is: less than equals is greater than Ffriction.
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vS=vL
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A small car of mass m and a large car of mass 4m drive along a highway at constant speed. They approach a curve of radius R. Both cars maintain the same acceleration a as they travel around the curve. How does the speed of the small car vS compare to the speed of the large car vL as they round the curve? vS=1/4vL vS=1/2vL vS=vL vS=2vL vS=4vL
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F1=1/3F2
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Now assume that two identical cars of mass m drive along a highway. One car approaches a curve of radius 2R at speed v. The second car approaches a curve of radius 6R at a speed of 3v. How does the magnitude F1 of the net force exerted on the first car compare to the magnitude F2 of the net force exerted on the second car? F1=1/3F2 F1=3/4F2 F1=F2 F1=3F2 F1=27F2