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course Phy 241
November 2 around 11:00am. I had some trouble with this lab. I attempted all, however. On the last question, I commented on what I didn't understand. Thanks for all your help!
`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?
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(a_up - a_down)/2=Ffric
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What therefore is the coefficient of friction?
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((a_up - a_down)/2)/mg=mu
ratio of forces is
m ((a_up - a_down)/2)/ (mg) = mu, so mu = ((a_up - a_down)/2)/ g, not / mg.
[since Fnet=Ffric=mu*mg]
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`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: 19.2, 22.8, 22.8
UP incline (vf=0): -35.3, -55.4, -107.65, -34.6
[What I did was take my counts, convert them to seconds (0.17sec for 1 count) and then use 40cm for DOWN the incline, which was length of incline and then used the measurements for UP the incline, which I measured the distance when it came to rest. I then used ‘ds=1/2a’dt^2, since v0=0, to solve for accelerations. The average of all the accelerations are in () at the end of each line. However, the UP accelerations are all diffe (DOWNWARD is POSITIVE)]
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`qx003. What therefore is the acceleration down, and what is the acceleration up?
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DOWN incline: 21.6cm/s^2
UP incline: -58.24cm/s^2
[I just took the average of the two sets.]
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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?
What is the direction of the frictional force as the car coasts up the incline?
Is the net force on the car greater when it coasts down the incline, or when it coasts up?
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Upward (opposite of motion)
Downward (opposite of motion)
Up the incline (there is more acceleration UP the incline, and therefore Fnet=ma, the Fnet will be greater)
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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?
• What is the expression for the net force on the car as it travels down the incline?
• What is the difference in the two expressions?
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Fnet= wt_parallel + f_fric (UP incline)
Fnet= wt_parallel - f_fric (DOWN incline)
The signs of f_fric, which friction is OPPOSITE of the motion.
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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?
• What is the magnitude of the net force on thc car as it travels up own the incline?
• What is the difference between the magnitudes of these two forces?
• What do you therefore conclude is the force due to friction?
• What would be the corresponding coefficient of friction?
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(120g)*(21.6cm/s^2)=2592 N
(120g)*(-58.24cm/s^2)=-6989 N
2592-(-6989)=9581 N
(-58.24-21.6)/2=-40 [2 Ffric=(a_up)-(a_down), Ffric=(a_up)-(a_down)/2]
Very good, but that would be 40 cm/s^2 / g, or about .04. See my previous notes.
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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?
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‘dPE=’dy*mg
‘dPE=mg_parallel*’ds
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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?
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Fnet=Ffric=mu*Fnorm=mu*mg
[‘dPE will be different]
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`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?
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1cm [y/20=0.05, 20*(0.05)=1]
‘dPE=(1cm)(120g)(980cm/s^2)=117600 [Using ‘dPE=’ds_y*mg] that's in g cm^2/s^2
I don’t’ know how to get the 3rd-5th question. What do I use to find the Ffric, the work friction did (W=F’dx?), and since KE=1/2mv^2, how do I get the velocity?
you have coeff of friction .04, normal force about equal in magnitude to weight, so you can find the frictional force; multiply it by the 20 cm displacement
‘dPE=mg_parallel*’ds ((120g)(980cm/s^2)(20cm)=2352000) What units would this be in? It’s a large number so should I have converted it to another unit?
multiply your units, which you included in the calculation; you get g cm^2 / s^2. Completely analogous to kg m^2 / s^2, the SI unit of work.
This unit is the erg.
The magnets will repel, energy is gained.
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Good work. See my questions and modify accordingly.