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course PHY 231
10/24 5:04 pm
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?
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The frictional force is related to the normal force which in turn is related to the grav_|| and so the a_up/down are related to g. How to relate the Ffr directly to a_up/down isn't clear to me at this point. a-up/down will both be down the incline and directly related to g_||, or the component of g parallel to the incline. So both Fric and Fgrav will be of the form (mg)_|| and µ(mg)_||.
you'll need to think about the net force in order to get the full insight into the situation
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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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1.75 s, 28 cm
I don't seem to have data for up the incline w/o the magnet system.
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`qx003. What therefore is the acceleration down, and what is the acceleration up?
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18.3 cm/s^2 (down)
we will repeat this using the TIMER; should be a quick repeat
instructions will be given
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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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Frict= up the incline
Frict= down the incline
Fnet > down incline b/c gravity_|| is working with the car' motion not against and Fgrav is >> Frict, mostly
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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_frict
Fnet= wt_parallel- f_frict
Difference is wt_parallel acting against then for the net force as it moves up then down the incline respectively
good, but it's probably more obvious what's going on if you don't change your positive direction
you would be wt_parallel - f_frict and wt_parallel + f_frict, which I think would make the difference in forces more direct and obvious
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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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Fnet= 120*18.3 = 2196 g cm/s^2 or 0.02 N
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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?
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`dPE_grav= -(mg) `dy
`dPE_el= +Fmag( `ds)
`dPE = -W so `dPE= (µmg) - (mg `dy)+Fmag( `ds)
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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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ATAN 0.05 ≈ 2.86ş, 20*sin 2.86 ≈ 1 cm
`dPE= -mg*1= -1.18 N*cm
µFnorm= µmg= 0.04 N
W_nc_on= -0.04*20= 0.8 N*cm
KE= 1.18 - 0.8= 0.38 N*cm
PE_el = 0.38 N*cm
Turns to KE_up at instant KE_down = 0 and is dissipated by W_cons_on + W_frict_on until the KE_up= 0. Each way friction dissipates the the total energy causing less and less movement which in turns leads to less PE/KE until the force of gravity down the ramp is balanced by the force of the magnet up it.
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&#Good responses. See my notes and let me know if you have questions. &#
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