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Various Laws of Motion, Energy, Momentum and Collision Problems

  1. Nov 22, 2003 #1
    I have this take home test I'm doing for Physics. I did all the problems and was just wondering if someone can correct me if I'm wrong on any of the problems because I'm on the borderline B-/C+ and I could really use your guys' help to check the problems. Thanks a lot.

    **=not sure/dont know how to do

    1) Two ropes are attached to a 40kg object. The first rope applies a force of 25N and the second 40N. If the two ropes are perpendicular to each other, what is the resultant acceleration of the object?

    Answer: Fnet= (25N^2 + 40N^2)^1/2 = 47.17N/40kg = 1.18m/s^2

    2) An airplane of mass 1.2 x 10^4kg tows a glider of mass .6 x 10^4kg. The airplane propellers provide a net forward thrust of 3.6 x 10^4N. What is the glider's acceleration?

    Answer: 3600N = (1200kg+600kg)a= 2m/s^2

    3)A baseball batter hits an incoming 40m/s fastball. The ball leaves the bat at 50m/s after a ball-on-bat contact time of 0.030s. What is the force exerted on the .15kg baseball?

    Answer: F= (.15kg)(90m/s) / .030s = 450N

    4)A 9.0kg hanging weight is connected by a string over a pulley to a 5.0kg block sliding on a flat table. If the coefficient of sliding friction is 0.20, find the tension in the string.

    Answer: a= (m2g-(.2)m1g)/(m1+m2) a= (9.0kg)(9.8m/s^2) - (0.20)(5.0kg)(9.8m/s^2) / (9.0kg + 5.0kg) = 5.6 m/s^2

    T= (5.0kg)(5.6m/s^2)+9.8N= 37.8

    5)A worker pulls a 200-N packing crate at constant velocity across a rough floor by exerting a force F = 55.0N at an angle of 35 degrees above the horizontal. What is the coefficient of kinetic friction of the floor?

    Answer: 55cos35N= mK (200N+55sin35N) = 45.05N/231.55N = mK = .195

    **6) A block is launched up an incline plane. After going up the plane, it slides back down to its starting position. The coefficient of friction between the block and the plane is 0.3. Compare the time for the trip up the plane with the time for the trip down the plane. Determine whether the ratio is greater than, equal to, or less than one.

    Answer: ????????????


    7) What is the minimum amount of energy required for an 80-kg climber, who is carrying a 20-kg pack to climb Mt Everest, 8,850m high?

    Answer: mgyi = (100kg)(9.8m/s^2)(8850m) = 8673kJ

    8) A 2.00kg ball has zero kinetic and potential energy. Ernie drops the ball into a 10.0m deep well.

    a) Determine the sum of its kinetic and potential energy just before the ball hits the bottom

    b) Determine the sum of its kinetic and potential energy after the ball comes to a stop in the mud.

    Answer: PE= 2kg x 9.8m/s^2 x -10m = -196J
    KE = Vi^2 + 2gh = 2(9.8m/s^2)(10m) = 196J

    196J-196J=0

    9) A parachutist of mass 50.0kg jumps out of an airplane at a height of 1000m. The parachute deploys, and she lands on the ground with a speed of 5.0m/s. How much energy was lost to air friction during this jump?

    Answer:KEf+PEf-KEi-PEi
    625J-0+0-49000J
    48375J lost.

    10) A 1000kg sports car accelerates from zero to 25m/s in 7.5s. What is the average power delivered by the automobile engine?

    Answer: KE = 1/2(1000kg)(25m/s)^2 = 312.5kJ
    P = KE/t = 312.5kJ/7.5s = 41.7kW

    11) A girl and her bike have a total mass of 40kg. At the top of the hill her speed is 5.0m/s. The hill is 10m high and 100m long. If the force of friction as she rides down the hill is 20N, what is her speed at the bottom?

    Answer: a= .475m/s^2
    10^2+100^2=x^2
    x=100.5m
    v=Vo^2+2ax
    v=(120.47m^2/s^2)^1/2 = 10.98m/s

    13) Popeye, of mass 70kg, has just downed a can of spinach. He accelerates quickly and stops Bluto, of mass 700kg, who is charging in at 10m/s. What was Popeye's speed?

    Answer: 700kg x 10m/s = 7000N 7000N/70kg=100m/s

    15) Object 1 has twice the mass of object 2. Each of the objects has the same magnitude of momentum. Determine the ratio of their kinetic energies.

    Answer: object 1:p=(10kg)(5m/s)
    object 2:p=(5kg)(10m/s)
    v of ob.2 = 2 times v of ob.1

    KE=1/2mv^2

    ob.1: 1/2(10kg)(5m/s)^2 = 125J
    ob.2: 1/2(5kg)(10m/s)^250J

    KE2 = 2KE1
     
    Last edited: Nov 24, 2003
  2. jcsd
  3. Nov 23, 2003 #2

    cepheid

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    In problem # 1 it says that the two ropes are perpendicular to each other, not antiparallel. Therefore, you cannot compute the resultant force merely by subtracting one force from the other. If you draw a free body diagram that represents the object as a point, you will see that the two force vectors form the arms of a right triangle whose hypotenuse is the resultant vector (in other words, the two forces add up to create a net force in a direction midway between the original two). The magnitude of this resultant force vector can be calculated using the Pythagorean theorem:

    [tex] \ F_{net} = \sqrt{(25N)^2 + (40N)^2} [/tex] = 47.17N

    You were correct that the resultant acceleration can be calculated from the resultant force using Newton's 2nd Law. Your resultant force just wasn't correct.
     
    Last edited: Nov 23, 2003
  4. Nov 23, 2003 #3

    cepheid

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    Your answer for problem # 3 is in joules, which is a unit of energy. But the question asks for the force exerted on the baseball by the bat. Your answer should be in newtons. Here is how I would go about solving many highschool physics problems. (Once you get to University, just do what I do; look at the problem, break down into tears, and go see the T.A. :wink: ):

    It often helps to assign a coordinate system to your problem. If we assume that the ball is in uniform motion in a straight line, then we can assign a 1D coordinate system to this problem in which velocities toward the batter are defined to be negative, and velocities away from the batter are positive.

    Now list the given information:

    The ball's initial and final velocities:

    [tex] \ v_i [/tex] = -40m/s
    [tex] \ v_f [/tex] = 50m/s

    The duration of the contact between the bat and the ball:

    [tex] \Delta t [/tex] = 0.030s

    The mass of the baseball [tex] m [/tex] = 0.15kg

    What quantity do we want to calculate? The force exerted by the bat on the ball: [tex] F [/tex] = ?

    I would calculate this force using the Impulse-Momentum Theorem, which states that the magnitude of the impulse exerted on an object is equal to the magnitude of the change in momentum:

    [tex] F \Delta t\ = \ m \Delta v [/tex]

    Solve for the force:

    [tex] F \ = \frac {m \Delta v}{\Delta t} [/tex]

    [tex] \Delta v \ =\ v_f - v_i [/tex] = 50m/s - (-40m/s) = 90m/s

    [tex] F [/tex] = (0.15kg)(90m/s) / (0.030s) = 450N

    The force is large because a large impulse (change in momentum) must be applied over a very short time interval. The force must accelerate the ball, sending it off in completely the opposite direction.
     
    Last edited: Nov 23, 2003
  5. Nov 23, 2003 #4

    cepheid

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    # 8: There is a major conceptual error here: when a conservative force does work on an object, the total mechanical energy (kinetic + potential) of the system remains constant. Since we start with zero energy, the sum of the kinetic and potential energy at the bottom of the well is still zero:

    The kinetic energy gained by the ball should equal the gravitational potential energy released by the ball/well system. Even though the potential energy of the system is initially zero, that is still much higher than it is when the ball is at the bottom of the well. The potential energy lost is: PE = mg(delta(h)) = (2.00kg)(9.81 N/kg)(-10.0m) = -196.2J

    The KE gained = 1/2mv^2, where v is the velocity at the bottom of the well.

    v^2 = ?

    v^2 = vi^2 + 2g(delta(h))

    = 2g(delta(h))

    = 2(9.81)(10.0m)

    therefore KE = 1/2(2kg)[2(9.81)(10.0m)] = 196.2J

    ETotal = KE + PE = 196.2J - 196.2J = 0

    I hope this helps. It's been a while since I've done kinematics, so I wouldn't mind if someone else double checked my work as well.

    I have to go and do my own HW, but I'll try some of these other questions later. Even if your take home test is already due, at least you'll know how to do the questions after the fact.

    Cheers
     
  6. Nov 24, 2003 #5
    I managed to figure out most of the problems! Thanks for the help.

    I'm still stuck on #6? Anyone have a clue to figure it out?
     
  7. Nov 24, 2003 #6

    Doc Al

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    The distance is the same for both trips. But the acceleration is not. Consider the net force on the block in each case. What's different?
     
  8. Nov 24, 2003 #7
    Fnet= mass x acceleration
    mass is the same. how do i get the acceleration going up and down the plane??
     
  9. Nov 25, 2003 #8

    Doc Al

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    Find the forces acting on the block. Add them up.
     
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