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course Phy 241
6/24 1:35 am. I do not understand question 12, I am suppose to find an exact numerical answer or guess whether the answer suppose twice as much or etc.
Question: `q001. Note that there are 15 questions in this set.
You have done the Introductory Force Experiment in which you used rubber bands and bags of water, and you understand that, at least in the vicinity of the Earth's surface, gravity exerts downward forces. You have also seen that forces can be measured in units called Newtons. However you were not given the meaning and definition of the Newton as a unit of force. You also know that the standard unit in which mass is measured is the kilogram.
Here we are going to develop, in terms of an experiment, the meaning of the Newton as a force unit.
We begin with a cart containing a number of masses.
• We suppose the cart contains 25 equal masses. We will call these the 'small masses'.
• The cart itself is equal in mass to the combined total of the 25 small masses (this can be verified by balancing them on a beam at equal distances from a fulcrum, the 25 small masses on one end and the cart on the other). So the entire system has a mass which is 50 times that of the small masses.
• The cart is placed on a slight downward incline and a weight hanger of negligible mass is attached to the cart by a light string and suspended over a low-friction pulley at the lower end of the ramp.
The incline is adjusted until the cart, when given a slight push in the direction of the pulley, is observed to move with unchanging, or constant, velocity (and therefore zero acceleration).
• At this slope the weight of the cart (which acts vertically downward), the force exerted by the incline in response to the weight (which acts perpendicular, or normal to the incline and is called the normal force) and the frictional force (which acts in the direction opposite motion) are all in balance. That is the sum of all these forces is zero.
• If any additional force is exerted in the direction of motion, that force will therefore be the net force.
The small masses are then moved one at a time from the cart to the hanger.
• In each case we regard the system as the cart and the suspended masses.
• Transfer of a mass from cart to hanger doesn't change the mass of the system, since the transferred mass is just moved from one part of the system to the other.
•
The gravitational, normal and frictional forces on cart and the masses that remain in it still add up to zero, for the same reasons as before.
However the suspended mass is no longer in the cart, and the force exerted on it by gravity is no longer balanced by the normal and frictional forces.
• The net force on the system is therefore now equal to the gravitational force on the suspended mass.
As masses are transferred one at a time, the system is therefore accelerated first by the force of gravity on one of the masses, then by the the gravitational force on two of the masses, etc..
• With the transfer of each mass we observer the time required for the system to accelerate from rest through a chosen displacement.
The acceleration of each system is then calculated, using the data for that system.
• The acceleration is graphed vs. the proportion of the total mass of the system which is suspended over the pulley.
Note that if the entire mass of the system, including the cart, is placed on the weight hanger, there will be no mass left on the incline and the entire weight will fall freely under the acceleration of gravity.
The proportion of the total mass of the system which is now on the hanger is calculated for each system.
• recall that the entire system is equivalent in mass to 50 of the small masses
• so for example if a single small mass is on the hanger, that corresponds to 1/50 the mass of the system and the proportion of the mass which is suspended is 1/50 = .02
• if two small masses are on the hanger the proportion is 2/50 = .04, etc.
Suppose the data points obtained for the second, fourth, sixth, eighth and tenth systems were
• (.04, 48 cm/s^2), (.08, 85 cm/s^2), (.12, 125 cm/s^2), (.16, 171 cm/s^2), (.20, 190 cm/s^2)
Sketch these points on an accurate graph of acceleration vs. proportion of weight suspended and determine the slope and y-intercept of the line.
• What is your slope and what is the y intercept?
• What is the equation of the line?
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Your solution:
The slope for the line of best fit is 925 and the y-intercept is 12.5.
The equation of the line of best fit is y = 925x +12.5
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Question: `q002. Do the points seem to be randomly scattered around the straight line or does there seem to be some nonlinearity in your results?
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Your solution:
Based on the four points, it seem like the points in the bottom is along the line of best fit while the points in the upper area seems to surround the line.
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Question: `q003. If the acceleration of the system is indeed proportional to the net force on the system, then your straight line should come close to the origin of your coordinate system. Is this the case?
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Your solution:
The acceleration of the system will be a constant multiple of the x-coordinate, in which, you can find the y-coordinate by multiplying the x-coordinate by a specific number. The specific number is the same for all x-coordinates.
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Question: `q004. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended?
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Your solution:
The frictional force and the force exerted by the ramp together counter the force of gravity on the cart and the masses remaining in it. The gravitational force on the cart and the masses in it therefore does not affect the acceleration of the system.
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Question: `q005. This results of this sort of experiment, done with good precision, support the contention that for a given mass the acceleration of a system is indeed proportional to the net force exerted on the system. Other experiments can be done using rubber bands, springs, fans and other nongravitational sources of force, to further confirm this result.
In another sort of experiment, we can see how much force is required on different masses to obtain a certain fixed acceleration. In such experiments we find for example that if the mass is doubled, it requires twice the force to achieve the same acceleration, and that in general the force required to achieve a given acceleration is proportional to the amount of mass being accelerated.
In a certain experiment using the same cart and masses as before, plus several additional identical carts, a single cart is accelerated by a single suspended mass and found to accelerate at 18 cm/s^2. Then a second cart is placed on top of the first and the two carts are accelerated by two suspended masses, achieving an acceleration of 20 cm / s^2. Then a third cart is placed on top of the first to and the three carts are accelerated by three suspended masses, achieving and acceleration of 19 cm/s^2. A fourth cart and a fourth suspended mass are added and an acceleration of 18 cm/s^2 is obtained. Adding a fifth cart in the fifth suspended mass an acceleration of 19 cm/s^2 is obtained. All these accelerations are rounded to the nearest cm/s^2, and all measurements are subject to small but significant errors in measurement.
How well do these results indicate that to achieve a given acceleration the amount of force necessary is in fact proportional to the amount of mass being accelerated?
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Your solution:
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Question: `q006. Now we note again that the force of gravity acts on the entire mass of the system when an entire system is simply released into free fall, and that this force results in an acceleration of 9.8 m/s^2. If we want our force unit to have the property that 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2, then how many force units does gravity exert on one mass unit?
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Your solution:
The amount of force units does gravity exert on one mass unit is 9.8 force units.
9.8 m/s^2 * 1 mass unit
= 9.8 force unit
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Question: `q007. If we call the force unit that accelerates 1 mass unit at 1 m/s^2 the Newton, then how many Newtons of force does gravity exert on one mass unit?
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Your solution:
The amount of Newtons does gravity exert on one mass unit is 9.8 Newtons.
9.8 m/s^2 * 1 mass unit
= 9.8 Newtons
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Question: `q008. The mass unit used here is the kilogram. How many Newtons of force does gravity exert on a 1 kg mass?
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Your solution:
The amount of Newtons does gravity exert on one kg is 9.8 Newtons
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Question: `q009. How much force would gravity exert on a mass of 8 kg?
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Your solution:
The amount of force or Newtons would gravity exert on a mass of 8 kg is 78.4 force units, or Newtons.
8kg * 9.8 m/s^2
= 78.4 kg*m/s^2 = Newtons
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Question: `q010. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2?
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Your solution:
The amount of force or Newtons would gravity exert on a mass of 5 kg at 4 m/s^2 is 20 force units or Newtons.
5 kg*4 m/s^2
=20 kg*m/s^2 =Newtons
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Question: `q011. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2?
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Your solution:
The amount of force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2 is 2400 Newtons.
1200 kg*2 m/s^2
= 2400 kg*m/s^2 = Newtons
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Question: `q012. A net force F_net accelerates a certain mass m at 30 cm/s^2.
What would be the acceleration of the same mass if subjected to a net force twice as great?
What would be the acceleration if the original net force acted on a mass half as great?
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Your solution:
The acceleration of the same mass if subjected to a net force twice as great would be also twice as much.
The acceleration if the original net force acted on a mass half as great would be half as much.
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Question: `q013. What net force would be required to accelerate a 50 kg mass at 4 m/s^2?
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Your solution:
The net force that would be required to accelerate a 50 kg mass at 4 m/s^2 is 200 Newtons.
50 kg*4 m/s^2
= 200 kg*m/s^2 = Newtons
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Question: `q014. What would be the acceleration of a 40 kg mass subjected to a net force of 20 Newtons?
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Your solution:
The acceleration of a 40 kg mass subjected to a net force of 20 Newtons is 2 m/s^2.
Newtons = kg*m/s^2
40 kg/20 kg*m/s^2
= 2 m/s^2
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Question: `q015. What is the mass of an object which when subjected to a net force of 100 Newtons accelerates at 4 m/s^2?
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Your solution:
The mass of an object which when subjected to a net force of 100 Newtons accelerates at 4 m/s^2 is 25 kg.
100 Newtons = kg*m/s^2 / 4 m/s^2
= 25 kg
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