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Physics problem 35
Problem 35




 An empty sled of mass 25 kg slides down a muddy
  hill with a constant speed of 2.4 m/s. The slope of the
  hill is inclined at an angle of 15º with the horizontal as
  shown in the figure above.
Part A
 Calculate the time it takes the sled to go 21 meters
  down the slope



                                   The velocity is constant so
                                   just divide 21 meters by 2.4
                                   meters per second.
Part A
         Meters cancel out and were
         left with a unit of time. That’s
         exactly what we want!
Part B
 On the dot below that represents the sled, draw and
  label a free-body diagram for the sled as it slides
  down the slope.
Part B




First, the weight force. Weight always points
straight down, even if the object is on a slope.
Part B




Now the normal force. The normal force is
always perpendicular to the surface that
the object is on.
Part B




Now friction. Friction is the along the
surface the object is traveling on.
Part C
 Calculate the frictional force on the sled as it slides
  down the slope.




                                    Change the axis so that the
                                    weight force is off at an angle
Part C
    Now that the FBD is rotated, the
    legs of the triangle formed by the
    weight force are equal to the
    Normal and friction forces
    because the sled is in equilibrium.
Part C
                 How to find the values of the normal
                 and friction forces. (aka the legs of
                 the triangle)


                                          Used 10 for
                                          gravity




MAKE SURE
YOUR
CALCULATOR IS
IN DEGREES!!!
Part C



          We use sine because friction equals the
          side opposite to the angle. After
250 N     multiplying by the weight force (the
          hypotenuse) we see that the force by
          friction is 64.705 Newtons
Part D
 Calculate the coefficient of friction between the sled
  and the muddy surface of the slope.




                                           To find the
                                           coefficient of
                                           friction we use the
                                           FUN equation.
Part D
Part D


Cosine 15 degrees times the weight
force equals the normal force (the
adjacent leg of the triangle)
                                     250 N
Part D
         Now we can plug in numbers to the
         FUN equation.



                   f = 64.705 N
                   N = 241.481 N
N        N


                   N
    N
    N          N           And solve for μ
Part E
 The sled reaches the bottom of the slope and
  continues on the horizontal ground. Assume the
  same coefficient of friction.
Part E
i.       In terms of velocity and acceleration, describe the
         motion of the sled as it travels on the horizontal
         ground.


     •     The acceleration of the sled is now
           negative, meaning the sled’s velocity is decreasing
           and it is slowing down.
Part E
ii. On the axes below, sketch a graph of speed (v)
    versus time (t) for the sled. Include both the sled’s
    travel down the slope and across the horizontal
    ground. Clearly indicate with the symbol tιthe time
    at which the sled leaves the slope.
Part E




The velocity is constant at 2.4 meters per second
until the sled leaves the slope. After that it begins
decreasing until velocity equals 0 and it stops
moving.

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Physics problem 35

  • 2. Problem 35  An empty sled of mass 25 kg slides down a muddy hill with a constant speed of 2.4 m/s. The slope of the hill is inclined at an angle of 15º with the horizontal as shown in the figure above.
  • 3. Part A  Calculate the time it takes the sled to go 21 meters down the slope The velocity is constant so just divide 21 meters by 2.4 meters per second.
  • 4. Part A Meters cancel out and were left with a unit of time. That’s exactly what we want!
  • 5. Part B  On the dot below that represents the sled, draw and label a free-body diagram for the sled as it slides down the slope.
  • 6. Part B First, the weight force. Weight always points straight down, even if the object is on a slope.
  • 7. Part B Now the normal force. The normal force is always perpendicular to the surface that the object is on.
  • 8. Part B Now friction. Friction is the along the surface the object is traveling on.
  • 9. Part C  Calculate the frictional force on the sled as it slides down the slope. Change the axis so that the weight force is off at an angle
  • 10. Part C Now that the FBD is rotated, the legs of the triangle formed by the weight force are equal to the Normal and friction forces because the sled is in equilibrium.
  • 11. Part C How to find the values of the normal and friction forces. (aka the legs of the triangle) Used 10 for gravity MAKE SURE YOUR CALCULATOR IS IN DEGREES!!!
  • 12. Part C We use sine because friction equals the side opposite to the angle. After 250 N multiplying by the weight force (the hypotenuse) we see that the force by friction is 64.705 Newtons
  • 13. Part D  Calculate the coefficient of friction between the sled and the muddy surface of the slope. To find the coefficient of friction we use the FUN equation.
  • 15. Part D Cosine 15 degrees times the weight force equals the normal force (the adjacent leg of the triangle) 250 N
  • 16. Part D Now we can plug in numbers to the FUN equation. f = 64.705 N N = 241.481 N N N N N N N And solve for μ
  • 17. Part E  The sled reaches the bottom of the slope and continues on the horizontal ground. Assume the same coefficient of friction.
  • 18. Part E i. In terms of velocity and acceleration, describe the motion of the sled as it travels on the horizontal ground. • The acceleration of the sled is now negative, meaning the sled’s velocity is decreasing and it is slowing down.
  • 19. Part E ii. On the axes below, sketch a graph of speed (v) versus time (t) for the sled. Include both the sled’s travel down the slope and across the horizontal ground. Clearly indicate with the symbol tιthe time at which the sled leaves the slope.
  • 20. Part E The velocity is constant at 2.4 meters per second until the sled leaves the slope. After that it begins decreasing until velocity equals 0 and it stops moving.