course Phy 201 ۪̀Βaoassignment #010
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19:46:14 Why does it make sense that the work done by gravity on a set of identical hanging washers should be proportional to the product of the number of washers and the distance through which they fall?
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RESPONSE --> The work done on an object is the force for or against that object. Gravity is doing work by acting upon the washers and pulling them down. On a greater slope, more washers would be necessary to move the cart at a constant velocity against gravity.
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19:47:12 ** When force is parallel to displacement, work is force * distance, so the proportionality with distance follows immediately. The force comes from the gravitational interaction of the masses of the washers with the Earth's gravitational field. More washers imply more force, and the force is proportional to the number of washers. **
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RESPONSE --> force is parallel to displacement, w=f*d more washers mean more force and force is proportional to number of washers
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19:48:56 Why is this consistent with the idea that the work done on a given cart on an incline is proportional to the vertical distance through which the cart is raised?
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RESPONSE --> the work done on the cart is the work to pull the cart up the incline at a constant velocity, so the cart is raised a certain vertical distance in relation to the number of washers used.
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19:49:12 ** The steeper the incline the more washers are required to pull the cart up the incline. In fact for small inclines and correcting for friction the number of washer is proportional to the slope of the incline. The vertical distance, being the rise through which the cart is raised, is also proportional to the slope of the incline. So for a given cart the number of washers is proportional to the incline. Since work is proportional to the number of washers we conclude the work is proportional to the vertical distance through which the cart is raised. **
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RESPONSE --> steeper incline needs more washers.
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19:50:33 How does the work done against friction of the cart-incline-pulley-washer system compare with the work done by gravity on the washers and the work done to raise the cart?
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RESPONSE --> total force is equal to the force of gravity minus the force of friction. when fnet is zero, the cart does not move. the forces acting on the cart are friction and gravity and the force to move the cart is the force of the washers.
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19:50:59 ** The work against friction plus the work done to raise the cart are all provided by the gravitational force acting on the descending weight. **
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RESPONSE --> Fg provides the frictional force and the work to raise cart.
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19:52:54 How do we establish, using a cart-pulley-washer system on a constant-velocity incline, that the acceleration of cart is proportional to the net force on the cart?
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RESPONSE --> We measure that the work done on the cart is proportional to the slope and then that the work done is proportional to the number of washers. this then translates to the work done on the cart is proportional to the net force on the cart.
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19:53:30 ** STUDENT ANSWER: In order to keep the velocity constant the amount of force should be constant but it must overcome the effects of gravity and therefore it must increase at a constant rate . The increase in the ramp slope will correspond to the increase in the force. INSTRUCTOR RESPONSE: Good start. This is half of it. This indicates how the force on the cart is proportional to acceleration. The other half is that the acceleration of the cart is proportional to the effective slope. Since the force and the acceleration of the cart are both proportional to the effective slope, they are proportional to each other. **
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RESPONSE --> the increase in slope is prop. to the increase in weights. this keeps the velocity constant.
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[ռFM䯟ԟ맗ٜeNzǶ assignment #011 㣾{WƁyࢌ Physics I Class Notes 10-19-2006
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19:55:25 How do the experiment with acceleration vs. number of washers and experiment with acceleration vs. ramp slope convince us that for a given cart there is a net force down the ramp which is proportional to the slope in excess of the constant-velocity slope (the slope excess)?
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RESPONSE --> The forces acting on a cart are the forces of the washers and the force of friction and gravity. when the slope is greater than the constant velocity slope, it becomes too much for the force of friction and the cart becomes susceptible to the force of gravity down the slope.
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19:56:27 ** A graph of acceleration vs. the number of washers in excess of the number required to maintain a constant velocity is consistent with a straight-line graph through the origin. **
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RESPONSE --> when the acceleration is compared to the number of washers used to exceed the constant-velocity slope, it is a linear progression of acceleration with slope increase.
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20:11:13 If for small slopes we find that the net force exerted on a cart is proportional to the slope excess, then why does Newton's Second Law ensure us that for small slopes the acceleration of the cart should in fact be proportional to the slope excess?
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RESPONSE --> If the slope is in excess of the constant velocity slope, then the amount of the acceleration will be in terms of the slope increase.
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20:11:41 ** Newton's Second Law tells us that for a fixed mass the acceleration is proportional to the force, so if for small slopes force is proportional to slope excess, then acceleration must also be proportional to slope excess. **
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RESPONSE --> acceleration is proportional to mass for a fixed mass
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20:12:18 How is the slope of the net force vs. acceleration graph for an object reveal the mass of the object? Why should a greater mass result in a greater slope?
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RESPONSE --> f=m*a: a greater mass will require a greater slope before the cart is a constant velocity
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20:12:32 ** Since F = m a, when we graph F vs. a the slope is m. **
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RESPONSE --> f vs. a, the slope is mass
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20:13:41 How do we obtain our definitions of work and kinetic energy in terms of the equation vf^2 = v0^2 - 2 a `ds and Newton's Second Law?
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RESPONSE --> we solve the equation for f`ds=1/2 (mvf^2-1/2mfo^2)
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20:14:05 ** Substituting a = Fnet / m into vf^2 = v0^2 + 2 a `ds we get vf^2 = v0^2 + 2 Fnet / m * `ds, which we easily rearrange to get Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2. Fnet `ds is defined as work, 1/2 m v^2 is defined as KE, and our equation tells us that the work done by the net force is equal to the change in KE. **
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RESPONSE --> fnet`ds is work=`KE
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