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`q001. For each of the given objects on the various inclines find the parallel and perpendicular components of the object's weight as a percent of its weight, using the sine and cosine functions. Compare with the estimates you made previously.
A car weighing 20 000 Newtons on an incline making angle 12 degrees with horizontal.
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Perpendicular = 20,000 * cos(258 degrees) = 4158.23 Newtons
Parallel = 20,000 * sin(258 degrees) = 19563 Newtons
@& The x axis is parallel to the incline.
The x component is 20 000 Newtons * cos(258 deg).
Review Introductory Problem Set 5, if necessary. *@
@& Looking ahead, you seem to be consistently reversing the meanings of 'parallel' and 'perpendicaular'.*@
#$&*A block of weight 30 pounds on a 37 degree incline.
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Perpendicular = 30 * cos(233) = 18.05 pounds
Parallel = 30 * sin(233) = 23.96 pounds
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`q002. Your answers to this question will be based on the results you got using the sines and cosines, and will differ from those you obtained previously using estimates. If the car in the first question experiences a frictional force which is 2% of the perpendicular component of its weight, then what is the magnitude of the frictional force?
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2% of 4158 newtons = 83.16 newtons
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If the car is coasting downhill what is the sum of the parallel component of its weight and the frictional force?
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19563 + (- 83.16) = 19479.8 newtons
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If the car is coasting uphill what is the sum of the parallel component of its weight and the frictional force?
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???
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@& If the car is coasting uphill what is the direction of the frictional force?
Is this force in the positive or negative direction of the appropriate axis?
How does this combine with the parallel component of the weight?
Be sure you've seen my note on 'parallel' and 'perpendicular'.*@
`q003. Again your results will be based on the calculations you did in the first question, and should be compared to your previous estaimtes.
If the hotwheels car in the first question is attached by a light thread to a washer weighing 20 000 dynes and suspended over a pulley at the lower end of the ramp, then if friction is ignored what is the net force acting in the direction down the ramp?
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27361.6 dynes + 20,000 dynes = 47361.6 dynes
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Answer the same question assuming that the washer is suspended from a pulley at the top of the ramp.
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20,000 dynes acting OPPOSITE weight. 27361.6 dynes + (- 20,000 dynes) = 7361.6 dynes
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`q004. Once more base your answers on the results obtained for the first question, and compare your results with your previous estimate-based results.
How much frictional force would it take to hold the block in the first question stationary on the incline?
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18.05 pounds
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What is this force as a percent of the weight of the block?
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18.05/30 = 60.16%
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`q005. Sound in air travels at about 340 m/s. If you drop a rock down a well which is 40 meters deep, how much time will elapse between the drop and hearing the sound of the splash?
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Distance: 40 meters, v0 = 0 m/s, acceleration = 9.8 m/s^2,
Time down: 2.02 seconds. Time sound to travel up: 0.117 seconds
Total time = 2.137 seconds
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`q006. A rock of mass 2 kilograms is tossed upward at 6 meters per second, being released at a height of 4 meters above the ground. Assume that air resistance has no significant effect on its motion.
How long after being released will it reach the ground?
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Velocity up = 6 m/s
Acceleration = -9.8 m/s^2 (gravity is acting in the opposite direction)
Time until velocity equals 0: 0.612 seconds
Distance travelled = 3 m/s (average velocity) * 0.612 seconds = 1.836 meters.
Distance above ground = 5.863 meters, time until hitting ground: s^2/9.8 m * 5.863 m = 0.598s^2 = 0.773 s
0.612 seconds + 0.773 seconds = 1.385 seconds
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@& Good.
Note you could have gotten the same result with one set of calculations.
v0 = +6 mm/s, a = -9.8 m/s^2, `ds = -4 m
should give you the same result.
Check it out.*@
What will be its kinetic energy at the instant it reaches the ground?
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9.8 m/s^2 * 2 kg = 19.6 newtons
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@& That's a force, not a kinetic energy.*@
What was its kinetic energy when just after its release?
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Just after release = 0 m/s to 6 m/s, acceleration = 6 m/s^2
6m/s^2 * 2 kg = 12 newtons
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@& That's not how you calculate KE.
And there's no quantity equal to 6 m/s^2. There is a quantity relevant to this question which is equal to 6 m/s, but not 6 m/s^2.*@
How much work did gravity do on the ball between release and striking the ground?
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9.8 m/s^2 * 2kg = 19.6 newtons of continuous force by gravity. 19.6 newtons * 1.385 seconds = 27.146 joules
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@& F_ave * `dt is impulse, which is equal to change in momentum.
But it's not work.*@
How much work did gravity do on the ball between release and reaching its maximum altitude?
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19.6 newtons * 0.612 seconds = 11.995 joules
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How much work did gravity do on the ball as it fell from its maximum altitude to the ground?
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19.6 newtons * 0.773 seconds = 15.15 joules
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`q007. A hotwheels car of mass 50 grams rolls down a ramp inclined at 30 degrees from horizontal.
How much force does gravity exert on it?
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9.8 m/s^2 * 50 grams = 490 g m/s^2 or 0.49 newtons
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@& A Newton is a kg m/s^2, not a g m / s^2.*@
What is the component of the gravitational force parallel to the incline?
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50 * sin(240 degrees) = 43.3 grams m/s^2 or 0.0433 newtons
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@& 50 presumably means 50 g.
Multiplying 50 g by the pure number sin(240 degrees) results in units of g, not units of g m/s^2.
43.3 g m/s^2 would indeed be .0433 Newtons, so it's a correctx conversion, but this isn't an answer to the given question.*@
If friction is negligible, what therefore will be the acceleration of the car?
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43.3 grams m/s^2 * 1/50 grams = 0.866 m/s^2
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`q008. A rubber band breaks when its tension reaches 20 Newtons. The rubber band is used to accelerate a 1.2 kg mass upward, with gradually increasing acceleration. When the acceleration reaches a certain value the rubber band breaks. What is that acceleration?
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`q009. Two dominoes are stacked, one on top of the other. The maximum frictional force exerted between the dominoes is 10% of the weight of the top domino. The bottom domino is attached to a rubber band chain, which is stretched to some length before the dominoes are released. This is repeated, slightly increasing the length of the rubber band chain with each repetition, until the top domino slides off.
What acceleration was necessary to cause this to occur?
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The acceleration must be as such that the force of the exerted on the bottom domino is exceeds the frictional force that keeps the two dominoes together.
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Will the top domino begin to slide shortly after the moment of release, or does this occur a bit later?
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The force exerted is greatest upon releasing the dominoes, so it will be shortly after the moment of release.
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`q010. A spring exerts an average force of 3 Newtons on a 100-gram toy car, initially at rest, as the car moves 10 centimeters. Other forces in the direction of the car's motion are not significant, nor is the mass of the spring.
How fast will the car be moving at that point?
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Mass = 100 grams, force = 3 newtons, or 3000grams m/s^2, distance = 10 cm
Acceleration = force/mass = 3000grams m/s^2 * 1/100grams = 30 m/s^2
V0 = 0, a = 30 m/s^ or 3000cm/s^2
Time spent = roughly 0.0577 seconds
@& You don't say how you got this.*@
0.0577 s * 30 m/s^2 = 1.731 m/s
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Assuming that the spring's force is conservative, by how much did its potential energy change during this interval?
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-3 newtons
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@& -3 Newtons is a force, not a change in PE.*@
`q011. A 100-gram steel ball observed to be moving at 80 cm/s collides with a marble of mass 30 kg, which is initially at rest. Immediately after the collision the ball is observed to be moving at 60 cm/s and the marble at 66 cm/s.
What is the total momentum of the two objects before collision?
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8000 gram cm/s
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What is their total momentum after collision?
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8000 gram cm/s
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What was the change in the momentum of the steel ball?
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30kg * 66 cm/s = 1980 kg cm/s
Initial momentum = 0
Change = 1980 kg cm/s
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What was the change in the momentum of the marble?
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Initial: 8000 gram cm/s
After collision: 6000 gram cm/s
-2000 gram cm/s
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How much total kinetic energy was present before the collision?
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unknown
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@& You have sufficient information, given at the beginning of the problem and not requiring anything you have calculated, to find the total KE both before and after collision.*@
How much total kinetic energy was present after the collision?
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Whatever the initial kinetic energy was (barring friction)
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@& Not necessarily so.
Energy is conserved, but KE is not necessarily conserved in a collision.
Momentum is conserved, as is energy, but KE is only part of the energy and energy can change form.*@
Estimate how long the steel ball and the marble would have been in contact.
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A split second
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@& From their speeds and diameters you can put an upper limit on how long they could have been in contact,
You can then revise this to a reasonable estimate.*@
Based on your estimate what average force was exerted by the ball on the marble, and what average force by the marble on the ball?
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`q012. An object moving at velocity v along the arc of a circle of radius r is being accelerated toward the center of the circle, with acceleration a_centripetal = v^2 / r. Note that 'centripetal' means 'toward the center'.
A domino of mass 15 grams is balanced each end of a metal strap of length 40 cm, which is rotating about its center at a rate that gives the domino a speed of 10 cm / second.
What is the centripetal acceleration of the domino?
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Acceleration = velocity^2/radius
(10cm/s)^2/40 cm
(100 cm^2/s^2)/40 cm = 2.5 cm/s^2
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@& The radius of the circle is not equal to the length of the strap.*@
What is the centripetal force on the domino?
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15 grams * 2.5 cm/s^2 = 37.5 gram cm/s^2
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Where does this centripetal force come from?
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The tension from the string, which is being held back by a singular point. This causes constant force to be applied, pulling the domino in a circle.
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How fast would the domino have to be moving in order for its centripetal acceleration to be 1.5 m/s^2?
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X^2/40 = 1.5
X^2 = 60, x = sqrt(60), roughly 7.746 m/s
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@& Using the correct radius of the domino's circular path you would have gotten a correct answer.*@
How fast would the domino have to be moving in order for the centripetal force to be 10% of its weight?
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10% of 15 grams = 1.5 grams
15 * (x^2/40) = 1.5, x^2/40 = 0.1
X^2 = 4
X = 2 m/s
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