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course Phy201
`q001. An incline is angled at 5 degrees above horizontal. A 15 kg block slides along the incline.If the xy axes are oriented so that the x axis is parallel to the incline, what are the x and y components of the block's weight?
Mgx = 15kg*9.8m/s^2cos(255) = -30.05
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-30.05 what? (i.e., include the units throughout, including with the end result).
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Mgy = 15kg*9.8m/s^2sin(255) = -141.99
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If the only forces acting perpendicular to the block are the y component of the weight and the normal force, then what is the normal force?
Fnormal = -mgy
Fnormal = 142N
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If friction exerts a force whose magnitude is equal to 8% of the magnitude of the normal force, then what is the magnitude of the frictional force?
142N*0.08 = 11.36N
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If the block is moving in the positive x direction, is the frictional force positive or negative (relative to the x direction)?
neg
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In this case what is the net force acting on the mass? (If you weren't able to figure out the frictional force, you can use 11 Newtons for the frictional force; that isn't quite right, so use the correct value if you got it.)
F =ma
F = 15kg(-9.8m/s^2)= -147N
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This is the gravitational force.
The net force is the sum of all the forces, including normal and frictional forces.
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What therefore is its acceleration?
a = Fnet/m
a = 147N / 15kg
a = -9.8m/s^2
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The object on the incline will not accelerate at 9.8 m/s^2.
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If the block is moving in the negative x direction, is the frictional force positive or negative (relative to the x direction)?
Pos…………..or would it be the same?
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In this case what is the net force acting on the mass?
Fnet = ma
Fnet = 147N
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What therefore is its acceleration?
9.8m/s^2
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You are not calculating net force; your acceleration on an incline is clearly less than 9.8 m/s^2.
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What definitions, laws and procedures did you use in answering these questions?
Laws of gravity and F=ma
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`q002. Suppose the block in the preceding slides 4 meters along the incline, in the positive x direction.
What work is done on it by the component of its weight parallel to the incline?
W = F*ds
W = 147N * 4m
W = -588J
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The component of the weight parallel to the incline is not 147 N.
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What work is done on it by the frictional force?
W = f*ds
W = 11.36N*4m
W45.44J
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You do not appear to have considered the signs of these quantities relative to the positive direction.
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What is the change in its gravitational PE, and how much work is done on it by nonconservative forces?
dPE = 588J
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The gravitational force and the displacement are not along the same line. You either need to use the component of the force along the line of motion, or the component of the displacement along the line of the force.
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What therefore is the change in its KE?
dKE = -588J
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Answer the same questions if the block slides 4 meters in the negative x direction.
dPE = -588J
dKE = 588J
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Is the magnitude of the KE change the same when the block slides in the positive direction as in the negative?
yes
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Why do you think this is?
Because it is a change in work (energy) nothing changes other than from a pos to neg direction.
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You need to carefully consider all displacements and forces relative to the positive x direction. There is a different in the net force between an object sliding up and an object sliding down an incline.
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Does the magnitude of the KE change depend on the initial velocity of the block?
yes
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What minimum KE would the block need to start with to travel at least 4 meters up the incline?
KE = 1/2mv^2
W = F*ds
W = 147N * 4m
It needs to be at least 588J
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What definitions, laws and procedures did you use in answering these questions?
Work energy theorem and F=ma and dW = dKE and dKE = -dPE
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`q003. Masses of 30 kg and 31 kg are suspended from a light frictionless pulley by a cord of negligible mass.
If the system starts from rest, and upon release accelerates through a 5 meter displacement, by how much does the gravitational PE of the less massive block change?
W = 294N *5m = 1470
dPE = -1470
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By how much does the gravitational PE of the more massive block change?
W = 303.8N * 5m = 1519J
dPE = -1519
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The PE change of a block will be negative if it moves downward.
However one block moves upward while the other moves downward, so you won't have a negative change in PE for both blocks.
Note that PE has units.
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What therefore is the change in the PE of the entire system?
Fnet = (mg1-mg2) = 49N
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The net force certainly could be m1 g - m2 g. However it's not clear how you got 49 N. The difference for these masses would be 9.8 N.
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W = 49N*5m = 245J
dPE = -245J
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What therefore is the change in the KE of the system?
dKE = 245J
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What therefore must be its KE at the end of the 5 meter displacement?
If it comes to rest then it is 0
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What is the mass of the system?
W = ms*ds
245 = m(9.8m/s^2) 5m
m= 5kg
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What therefore must be its velocity at the end of the 5 meter displacement?
0 because it comes to rest
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There is nothing mentioned in the problem that would bring the system to rest. The system moves freely through the entire 5 meter displacement.
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What is the net force acting on the system?
Fnet = mg1-mg2 = 49N
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What therefore is the acceleration of the system?
a = F/m
a = 49N/5kg = 9.8m/s^2
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The system contains masses of 30 kg and 31 kg. There is no 5 kg mass involved.
Two masses suspended over a pulley will clearly have less acceleration than if the masses were both in free fall (as they would be, for example, if the string was cut).
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What velocity would the system attain with this acceleration, starting with initial velocity 0 and accelerating through the 5 meter displacement?
KE = 1/2mv^2
245 = ½(5kg)(v^2)
a =9.8m/s
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What definitions, laws and procedures did you use in answering these questions?
Work energy theorem and F =ma
W = F*ds
KE = 1/2mv^2
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`q004. For the system of the preceding, suppose the 30 kg mass is placed on a frictionless horizontal surface, with the 31 kg mass suspended over the pulley. If the system is released from rest and accelerates through a 5 meter displacement:
By how much would its gravitational PE change?
F = ma
F = 30kg(9.8m/s^2) = 294N
F = 31kg(9.8m/s^2) = 303.8N
W = F*ds
W = 9.8N(5m) = 49J
dW = dKE
dKE = -dPE
dPE = -49N
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By how much would its KE change?
dKE = 49N
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What velocity would it therefore attain?
KE = 1/2mv^2
49J = ½(1kg)(v^2)
V = 9.8m/s
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What would be its acceleration?
9.8m/s^2 is this wrong I keep getting the same answer???????????
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Again you seem to be invoking a 5 kg mass, which is not part of this situation.
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Assuming this acceleration, what would be the velocity at the end of the 5 meter displacement?
V = 9.8m/s
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What definitions, laws and procedures did you use in answering these questions?
Work energy theorem and KE = 1/2mv^2
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`q005. University Physics only:
Report your data from today's experiment, including a brief description, your analysis and your conclusions.
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You're going to want to revise much of your work on this exercise. Check my notes.
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