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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=15kgx9.8m/s^2cos255=-30.05
Mgy=15kgx9.8m/s^2sin255=-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?
142x.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)?
Friction is negative
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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
Fnet=15(-9.8)=-147N
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This is the weight of the block, not the net force acting on it. The normal force counters most of the weight.
The block doesn't accelerate in the y direction, so the net force is in the x direction. To get the net force you have to add up all the forces in the x direction.
What forces act in the x direction?
What is their sum?
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What therefore is its acceleration?
A=fnet/m
A=147N/15kg
A=-9.8m/s^2
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This can't be so. It's clear that a block sliding along an incline accelerates at a lesser rate (much lesser) than an object dropped and allowed to accelerate freely.
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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)?
Friction would be positive, or the same
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In this case what is the net force acting on the mass?
Fnet=ma
147N
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What therefore is its acceleration?
+9.8m/s^2
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again the net force is not equal to the weight, and the acceleration is much less than the acceleration of gravity
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What definitions, laws and procedures did you use in answering these questions?
Laws of gravity
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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=Fxds
W=147Nx4m
W=588J
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The 147 N weight doesn't act in the direction of motion. A component of it does, but only a component and in this case, a relatively small component.
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What work is done on it by the frictional force?
W=fxds
W=11.36nx4m
W=45.44J
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this would be correct
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What is the change in its gravitational PE, and how much work is done on it by nonconservative forces?
PE=-KE
PE=588J
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The block slides 4 m along the incline, it doesn't move 4 meters vertically.
588 J would be magnitude of the PE change of the object if it moved 4 meters vertically.
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What therefore is the change in its KE?
dKE=-588
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Both the work done by friction and the work done by gravity must be taken into account.
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Answer the same questions if the block slides 4 meters in the negative x direction.
PE=-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
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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=fxds
147x4m
588J
It needs to be atleast 588J
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What definitions, laws and procedures did you use in answering these questions?
Work energy theorem
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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=294x5m=1470J
dPE=-1470
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By how much does the gravitational PE of the more massive block change?
F=31(9.8m/s^2)
W=303.8x5m=1519
dPE=-1519
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You give both PE changes as negative.
If one block falls, the other rises. The signs of the PE changes won't both be the same.
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What therefore is the change in the PE of the entire system?
Fnet=mg1-mg2=49N
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You're right that it's mg2 - mg1, but the difference isn't 49 N. You previously got gravitational forces 294 N and 303.8 N; from these quantities you can find the correct net force.
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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?
KE=+245J
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What therefore must be its KE at the end of the 5 meter displacement?
If it stops then KE=0
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There is no assumption that the system stops, and no reason to assume this. The system moves freely, and at the end of the 5 m displacement it has acculumated the KE you will have previously calculated.
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What is the mass of the system?
245=mx9.8m/s^2x5
M=5kg
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Neither of the masses is as little as 5 kg.
The net force on the system isn't equal to the weight of either object.
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What therefore must be its velocity at the end of the 5 meter displacement?
None because it comes to rest
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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=49N/5kg
A=9.8m/s^2
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Clearly this nearly counterbalanced system will not accelerate at anything close to the acceleration of gravity.
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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?
245=1/2/95kg)(v)^2
V=9.8m/s
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What definitions, laws and procedures did you use in answering these questions?
F=ma
W=fxds
dKE=w
dKE=-PE
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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?
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
49=1/2(1)v^2
V=9.8m/s
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What would be its acceleration?
A=9.8m/s^2
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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
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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 doing a lot of things right, but you're missing some of the important points.
You should rework this and submit the revision.
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