How Do Friction and Tension Affect Motion in Physics Problems?

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SUMMARY

This discussion focuses on the physics concepts of friction and tension as they apply to motion problems involving a hockey puck and a train. The first question involves calculating the distance a puck slides on ice with a kinetic friction coefficient of 0.09 after being shot at 25 m/s. The correct distance is determined using the work-energy principle, yielding 354 meters. The second question addresses the tension forces between train cars and the braking force when the train decelerates from 20 m/s to a stop over 23 meters, with calculations involving mass, acceleration, and friction.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the work-energy principle
  • Knowledge of kinetic friction coefficients
  • Ability to apply kinematic equations in physics
NEXT STEPS
  • Study the work-energy theorem in detail
  • Learn to calculate tension forces in multi-body systems
  • Explore the effects of varying friction coefficients on motion
  • Practice solving complex kinematic problems involving deceleration
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of friction and tension in motion problems.

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Homework Statement


q1: A hockey player shoots a 200 gram puck at 25m/s. the ice has a friction of 0.09 kinetic.
how far will the puck slide before it comes to a rest?

q2: A train has 3 cars, mc, mb, and ma. The acceleration of the train is 1.54 m/s^2. ma has a mass for 10kg, mb a mass of 35kg, mc has a mass of 20kg.The force of the train is 100 N forward.
Note: No friction in part a and b!
<mc-mb-ma> <-train going east
a) Calculate the tension force between ma and mb as well as the tension force between mb and mc.
b)Also calculate the braking force of the train if it is moving at 20 m/s when the brakes are suddenly applied. if it takes 23m to stop, calculate the braking force.
c) (with friction) If the coefficient of kinetic friction is equal to 0.10, calculate the engine force required to keep the train moving at a constant velocity. there are only frictional forces on car b and c.
d)Calculate the engine force that is required to accelerate the train at a rate of 1.0m/s^2[E]

Homework Equations


d=vf^2 - vi^2 / 2a(i think?)
f=ma
a=fnet/mass

The Attempt at a Solution


For q2 I'm not too sure about. I did 45(35+10) * 1.54 for ma-mb, and the same format for the next part. both answers, however, are wrong. can someone please help me through the steps?
For q1, i figured out the normal force was 1962 (200g * 9.81). The Force of friction would be 176.58 (1962 x 0.09) -25 since Fn and Fg cancel out. my acceleration from this was 151.58/200=0.7579
i put it into the kinematic eq d=vf^2 - vi^2 / 2a and got 236.84m, or 200m with significant digits. The correct answer is 354m. I am not sure how they got that answer, please explain?
 
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For the first question remember that the work done is equal to the change in kinetic energy, friction is the only force doing any work which means...
<br /> mg\mu x = \frac{1}{2}mv^2<br />
Where m is the mass of the puck, g is the acceleration of gravity on earth, \mu is the coefficient of friction, x is the distance traveled and v is the initial velocity of your puck.
 
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