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course PHY 201
10/25 9pm On this assignment, I wanted to submit what I already did and there are questions inside. Thanks.
Lab-related questions`qx001. Suppose a car of mass m coasts along a slight constant incline.
If the magnitudes of its accelerations while traveling up, and down, the incline are respectively a_up and a_down, then what is the magnitude of the frictional force acting on it?
What therefore is the coefficient of friction?
`qx002. Give your data for the acceleration of the car down the incline, and its acceleration up the incline:
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Down incline:
The height of the ramp is 2cm and the length is 40cm. Therefore, the slope is .05
1st trial `ds=40cm `dt=1.2s (8 counts)
2nd trial `ds=40cm `dt=1.2s (8 counts)
3rd trial `ds=40cm `dt=1.2s (8 counts)
4th trial `ds=40cm `dt=1.2s (8 counts)
Up incline:
1st trial `ds=14cm `dt= .6s (4 counts)
2nd trial `ds=19cm `dt= .6s (4 counts)
3rd trial `ds=32cm `dt= .9s (6 counts)
4th trial `ds=17cm `dt= .6s (4 counts)
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`qx003. What therefore is the acceleration down, and what is the acceleration up?
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Down incline:
V0=0
`ds=40cm
`dt=1.2s
Vave=33.3cm/s
VF=66.6cm/s
A=(66.6cm/s)/1.2s=55.5cm/s^2
Up incline:
1st trial
`ds=14cm
`dt= .6s
v0=0
vF=0
vAVE=23.3cm/s
a= (-23.3cm/s)/.6s= -38.8cm/s^2
2nd trial
`ds=19cm
`dt= .6s (4 counts)
v0=0
vF=0
vAVE=31.7cm/s
a= (-31.7cm/s)/.6s=-52.8cm/s^2
3rd trial
`ds=32cm
`dt= .9s
v0=0
vF=0
vAVE=35.6cm/s
a=`dv/`dt=(-35.6cm/s)/.9s= -39.6cm/s^2
4th trial
`ds=17cm
`dt= .6s
v0=0
vF=0
vAVE=28.3cm/s
a=-28.3cm/s/.6s= -47.2cm/s^2
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`qx004. The only forces acting on the coasting care are the component of its weight parallel to the incline, and the frictional force. The former is always directed down the incline. Let the direction down the incline be chosen as the positive direction.
What is the direction of the frictional force as the car coasts down the incline?
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Frictional force is in the opposite direction (up the incline)
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What is the direction of the frictional force as the car coasts up the incline?
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Frictional force is in the opposite direction (down the incline)
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Is the net force on the car greater when it coasts down the incline, or when it coasts up?
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Net force is greater going up because you are going against the direction of gravity
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`qx005. This question asks for symbols, but the expressions are short and everyone should answer this question:
Let wt_parallel stand for the parallel component of the weight and f_frict for the magnitude of the frictional force.
· What is the expression for the net force on the car as it travels up the incline?
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Fnet= wt_parallel*f_frict
you don't multiply two forces to get a net force, you add them (with appropriate signs)
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· What is the expression for the net force on the car as it travels down the incline?
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Would this and the following question not be the same?
Fnet= wt_parallel*f_frict
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· What is the difference in the two expressions?
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Fnet= wt_parallel*f_frict
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`qx006. If the mass of your car and its load is 120 grams, then based on your calculated accelerations:
· What is the magnitude of the net force on the car as it travels down the incline?
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Fnet=m*a=120g*55.5cm/s^2=.12kg*.555m/s^2=.07N
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· What is the magnitude of the net force on the car as it travels up the incline?
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Fnet=m*a=120g*38.8cm/s^2=.12kg*.388m/s^2=.05N
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· What is the difference between the magnitudes of these two forces?
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.02N
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· What do you therefore conclude is the force due to friction?
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.02N
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· What would be the corresponding coefficient of friction?
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Using the formula coefficient of friction=accel of car/accel of gravity, but I don’t think I use this formula because I would get 2 coefficients of friction.
Friction isn't the only force acting in the direction of the car's motion. If it was, then your formula would be correct (it would apply, for example, to a car coasting on a level surface).
Since the car is coasting on an incline, the component of the gravitational force parallel to the incline is not zero, and does act along the line of motion.
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`qx007. A car and magnet, with total mass m, coasts down an incline. A magnet at distance L from the position of release brings the car to rest (just for an instant) after it has coasted a distance `ds, during which its vertical position decreases by distance `dy. If energy lost to friction is negligible, then
· What is the change in the car's gravitational potential energy from release to the instant of rest?
· What is the change in the car's magnetic potential energy between these two points?
If the incline is slight, then the normal force on the car is very nearly equal and opposite its weight. If the coefficient of friction is mu, then how do your answers to the above two questions change?
`qx008. Assume your car, which has mass 120 grams, coasted 20 cm down an incline of slope .05, and that the coefficient of friction was .03. At the end of the incline, 25 cm from the initial position of the car, is a magnet, which brings the car to rest for an instant, at the end of its 20 cm displacement.
· How far did the car descend in the vertical direction, based on the 20 cm displacement and the slope .05?
· By how much did its gravitational PE therefore change?
· Assuming that the normal force is very nearly equal to the car's weight, what was the frictional force on the car?
· How much work did friction therefore do on the car?
· In the absence of the repelling magnet, how much KE would you therefore expect the car to have at the end of the 20 cm?
· How much magnetic PE do you therefore think is present at the instant of rest?
· What do you think will happen next to this magnetic PE?
Much of your work is good, but there are some errors. See my notes. You should submit a revision to be sure you are clear on these ideas.
&#Please see my notes and submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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