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What is the magnitude of the net force acting on the ball?

  1. Aug 3, 2009 #1
    1)When a 58 g tennis ball is served, it accelerates from rest to a speed of 45 m/s. The impact with the racket gives the ball a constant acceleration over a distance of 44 cm. What is the magnitude of the net force acting on the ball?

    2)A skater with an initial speed of 7.60 m/s is gliding across the ice. Air resistance is negligible. (a) The coefficient of kinetic friction between the ice and the skate blades is 0.100. Find the deceleration casued by kinetic friction. (b) How far will the skater travel before coming to rest?

    Forces and Newton's Laws of Motion
    please give me a given and formula

    answer at the back book
    1) 130N
    2) a) 0.980 b) 29.5 m
     
    Last edited: Aug 3, 2009
  2. jcsd
  3. Aug 3, 2009 #2
    attempt the problems first then tell me which part u having trouble with
     
  4. Aug 3, 2009 #3
    Forces and Newton's Laws of Motion
    please give me a given and formula

    answer at the back book
    1) 130N
    2) a) 0.980 b) 29.5 m
     
  5. Aug 3, 2009 #4
    thats not really attempting the problem

    here are a few hints:

    question 1) find the acceleration and multiply it by the mass
    2) do they tell u the mass of the sk8er? u need the mass to work out the kinetic friction
     
  6. Aug 3, 2009 #5
    nevermind, for question 2 u dont need the mass:

    calculate decceleration: final velocity squared= initial velocity squared + 2as
    work done = change in kinetic energy = force x distance the mass will cancel out
     
  7. Aug 3, 2009 #6
    tnx for your helping me
     
  8. Aug 3, 2009 #7
    These are both questions dealing with work and energy. A kinematic perspective will suffice as well, since we're dealing with constant accelerations, but the understanding you should be striving for is one of work and energy.

    Consider the initial and final energies for both cases, and consider the forces performing work on the tennis ball, and skier, respectively.

    For the tennis ball, the work is the average net force, times the distance over which it acted. (Remember proper unit conversions!)

    For the skater, draw an FBD, and remember that rearranging Newton's second law provides us with: [tex]\vec a = \frac{\vec F}{m}[/tex]
    And again remember that the work of the net force is equal to the change in kinetic energy (Work-energy theorem) [tex]W= \bar Fd=\Delta E_k[/tex]
     
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