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course Phy 241
6/28 7:30 pm
011. Note that there are 12 questions in this set.
.Situations involving forces and accelerations.
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Question: `q001. A cart on a level, frictionless surface contains ten masses, each of mass 2 kg. The cart itself has a mass of 30 kg. A light weight hanger is attached to the cart by a light but strong rope and suspended over the light frictionless pulley at the end of the level surface.
Ignoring the weights of the rope, hanger and pulley, what will be the acceleration of the cart if one of the 2 kg masses is transferred from the cart to the hanger?
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Your solution:
m =2 kg, a = 9.8 m/s^2, m2 = 30 kg, m1 = 10 masses
2 kg * 9.8 m/s^2
= 19.6 Newtons
(Ten masses * 2 kg) + 30 kg
= 20 kg + 30 kg
= 50 kg
19.6 Newtons/50 kg
= .392 m/s^2
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Question: `q002. Two 1-kg masses are suspended over a pulley, one on either side. A 100-gram mass is added to the 1-kg mass on one side of the pulley. How much force does gravity exert on each side of the pulley, and what is the net force acting on the entire 2.1 kg system?
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Your solution:
100 g = .1 kg
1 kg * 9.8 m/s^2
= 9.8 Newtons
(.1 kg + 1 kg) * 9.8 m/s^2
= 1.1 kg * 9.8 m/s^2
= 10.78 Newtons
(10.78 Newtons - 9.8 Newtons)/ 2.1 kg
= .98 Newtons/2.1 kg
= .45 m/s^2
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Question: `q003. If in the previous question there is friction in the pulley, as there must be in any real-world pulley, the system in the previous problem will not accelerate at the rate calculated there. Suppose that the pulley exerts a retarding frictional force on the system which is equal in magnitude to 1% of the weight of the system. In this case what will be the acceleration of the system, assuming that it is moving in the positive direction (as defined in the previous exercise)?
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Your solution:
2.1 kg * 9.8 m/s^2
= 20.58 Newtons
1% * 20.58 Newtons
.01 * 20.58 Newtons
=.2058 Newtons
Opposed force = -.21 Newtons
10.78 Newtons - 9.8 Newtons - 2.1 Newtons
= .77 Newtons
.77 Newtons/(2.1 kg)
= .37 m/s^2
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Question: `q004. Why was it necessary in the previous version of the exercise to specify that the system was moving in the direction of the 1.1 kg mass. Doesn't the system have to move in that direction?
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Your solution:
If the system is released from rest, since acceleration is in the direction of the 1.1 kg mass its velocity will certainly always be positive.
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Question: `q005. If the system of the preceding series of exercises is initially moving in the negative direction, then including friction in the calculation what is its acceleration?
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Your solution:
Fnet = 10.78 Newtons - 9.8 Newtons + .21 Newtons
Fnet = 1.19 Newtons
a = 1.19 Newtons/(2.1 kg)
a = .57 m/s^2
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Question: `q006. If friction is neglected, what will be the result of adding 100 grams to a similar system which originally consists of two 10-kg masses, rather than the two 1-kg masses in the previous examples?
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Your solution:
10 kg to 10.1 kg
(10.1 kg * 9.8 m/s^2) - (10 kg * 9.8 m/s^2)
= 98.98 Newtons - 98 Newtons
= . 98 Newtons
10.1 kg + 10 kg
= 20.1 kg
a = .98 Newtons/20.1 kg
a = .048 m/s^2
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Question: `q007. If friction is not neglected, what will be the result for the system with the two 10-kg masses with .1 kg added to one side? Note that by following what has gone before you could, with no error and through no fault of your own, possibly get an absurd result here, which will be repeated in the explanation then resolved at the end of the explanation.
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Your solution:
0.01 * 20.1 kg * 9.8 m/s^2
= 1.97 Newtons
98.98 Newtons - 98 Newtons - 1.97 Newtons
= -.99 Newtons
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Question: `q008. A cart on an incline is subject to the force of gravity. Depending on the incline, some of the force of gravity is balanced by the incline. On a horizontal surface, the force of gravity is completely balanced by the upward force exerted by the incline. If the incline has a nonzero slope, the gravitational force (i.e., the weight of the object) can be thought of as having two components, one parallel to the incline and one perpendicular to the incline. The incline exerts a force perpendicular to itself, and thereby balances the weight component perpendicular to the incline. The weight component parallel to the incline is not balanced, and tends to accelerate the object down the incline. Frictional forces tend to resist this parallel component of the weight and reduce or eliminate the acceleration.
A complete analysis of these forces is best done using the techniques of vectors, which will be encountered later in the course. For now you can safely assume that for small slopes (less than .1) the component of the gravitational force parallel to the incline is very close to the product of the slope and the weight of the object. [If you remember your trigonometry you might note that the exact value of the parallel weight component is the product of the weight and the sine of the angle of the incline, that for small angles the sine of the angle is equal to the tangent of the angle, and that the tangent of the angle of the incline is the slope. The product of slope and weight is therefore a good approximation for small angles or small slopes.]
What therefore would be the component of the gravitational force acting parallel to an incline with slope .07 on a cart of mass 3 kg?
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Your solution:
Gravitational force
(3 kg * 9.8 m/s) * .07
= 29.4 Newtons * .07
= 2.05 Newtons = 2.1 Newtons
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Question: `q009. What will be the acceleration of the cart in the previous example, assuming that it is free to accelerate down the incline and that frictional forces are negligible?
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Your solution:
a = 2.1 Newtons/(3 kg)
a = .7 m/s^2
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Question: `q010. What would be the acceleration of the cart in the previous example if friction exerted a force equal to 2% of the weight of the cart, assuming that the cart is moving down the incline? [Note that friction is in fact a percent of the perpendicular force exerted by the incline; however for small slopes the perpendicular force is very close to the total weight of the object].
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Your solution:
2% = .02
.02 * 29.4 Newtons
= .59 Newtons
Fnet = 2.1 Newtons - .59 Newtons
Fnet = 1.5 Newtons
a = Fmet/m
a =1.5 Newtons/3 kg
a =.5 m/s^2
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Question: `q011. Given the conditions of the previous question, what would be the acceleration of the cart if it was moving up the incline?
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Your solution:
Fnet = 2.1 Newtons +.59 Newtons
Fnet = 2.69 Newtons = 2.7 Newtons
a = 2.7 Newtons/3 kg
a = .9 m/s^2
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Question: `q012. Assuming a very long incline, describe the motion of a cart which is given an initial velocity up the incline from a point a few meters up from the lower end of the incline. Be sure to include a description of any acceleration experienced by the cart.
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Your solution:
The cart begins with a velocity up the incline, which we still taken to be the negative direction, and an acceleration of +.9 m/s^2. its velocity will later be 0 for an instant, immediately after which it begins moving down the incline as result of the acceleration provided by the weight component parallel to the incline. As soon as it starts moving down the incline its acceleration decreases to +.5 m/s^2, but since the acceleration and velocity are now parallel the cart speeds up, increasing its velocity by .5 m/s every second, until it reaches the lower end of the incline
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