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course Phy 241
10/24 8
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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Kentic friction force = (mu sub k)*normal force, where mu sub k is coefficient of friction
your answer would be in terms of a_up and a_down
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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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Knowing there is accel we can use Newtons 2nd law, Fnet = m*accel(vector)
For a_down
y-component of accel a sub y is zero, but x-component is not a sub x
Sum of x forces = mg*sin(`theta) + (-Ffrict) = m*a sub x
Sum of y forces = n + (-mg*cos(`theta)) = 0
Using second equation
n = mg*cos(`theta), which we can solve for Ffrict
Ffrict = mu sub k*n = mu sub k * mg*cos(`theta)
Now we su this expression back into x-component equation
mg*sin(`theta) + (-mu sub k * mg*cos(`theta)) = m*a sub x
So that when we solve for accel we get
a sub x = g( sin(`theta) mu sub k* cos(`theta)
For a_up
Solution is very similar the signs would be slightly different.
Looking at it symbolicly you can see
If for a_down, Fnet = m*a(vector)+Ffrict.
Then for a_up, Fnet must = m*a(vector)-Ffrict(friction force is now down ramp)
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qx003. What therefore is the acceleration down, and what is the acceleration up?
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For Fnet = m*a(vector)+Ffrict, for a-down
Then a_down = Fnet/m = [m*a(vector)+Ffrict]/m = a(vector)+Ffrict
For a_up = a(vector)-Ffrict
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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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Neg, Positive. I would think that Fnet is greater when init velocity is down.
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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, when moving up incline
Fnet = wt_parallel + f_frict, when moving down incline
Difference is when moving up incline Fnet is wt_parallel f_frict(frictional force is subtracted) but when moving down incline Fnet is wt_parallel + f_frict(frictional force is added)
if you subtract the down from the up, then, you get 2 f_frict, as I see you did below
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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 = 1.176Nsin(`theta) + f_frict,
Fnet = 1.176Nsin(`theta) - f_frict,
2* f_frict, exactly
??????,
coefficient of friction = mu_k*n = mu_k*1.176Ncos(`theta)
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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 = `ds*(mg*cos(`theta))
`dPE_mag = `ds*(mg*sin(`theta))
The f_frict force will make the change in PE for grar smaller as well as for mag.
Theta isn't given. You can calculate directly from the given quantities:
From position of release to position of rest:
`dPE_grav = - m g `dy
`dKE = 0 (v is zero at both points)
`dW_noncons = 0 (frictionless, no other nonconservative forces parallel to incline)
so `dPE = 0
we conclude that `dPE_mag = - `dPE_grav
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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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Vert dist travled is approx 1cm, found by 20cm*sin(.05) = .9996cm,
`dPE_grav = `ds*mg_paralle = 20cm*1.176N*cos(270) = 23.1527N, N * cm, not just N
Fnormal = 1.176N*sin(270) = approx -.207N, |f_frict| = .03*-.207N = .00621N
.00621N,
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Very good work. See my notes.
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