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A team of dogs accelerates a 290kg dogsled from 0 to 6.0m/s in 3.0 s. Assume that the acceleration is constant.

Part A
What is the magnitude of the force exerted by the dogs on the sled?
Part B
What is the work done by the dogs on the sled in the 3.0 s?
Part C
What is the instantaneous power of the dogs at the end of the 3.0 s?
Part D
What is their instantaneous power at 1.5 s?

Answer :

cjrodriguez89

Answer:

A) 580 N B) 5,220 J C) 1,740 W D) 870W

Explanation:

A)

  • Assuming no friction present (or that is negligible) the force exerted by the  dogs on the sled must meet Newton's 2nd law:

       [tex]F = m*a (1)[/tex]

  • We can find the value of the acceleration (assumed to be constant), just applying the definition of acceleration, as follows:

        [tex]a =\frac{v_{}-v_{0}}{t-t_{0}}[/tex]

  • where v = 6.0 m/s, v₀=0, t(0)=0, t=3.0s
  • Replacing by the values above, and solving for a:

        [tex]a =\frac{6.0m/s}{3.0s} = 2 m/s2[/tex]

  • Replacing this value of a , and m= 290 kg, in (1), we get:

       [tex]F = m*a = 290 kg*2 m/s2 = 580 N[/tex]

  • The magnitude of the force exerted by the dogs on the sled is 580 N.

B)

  • In order to be able to find the work done during the 3.0 s, we need to find the displacement produced by the force during that time.
  • As the acceleration is assumed to be constant, and the sled starts from rest, we can use the following kinematic equation:

        [tex]x = \frac{1}{2} * a * t^{2}[/tex]

  • Replacing by the values of a and t, we can find the displacement x (assuming x₀ = 0), as follows:

       [tex]x = \frac{1}{2} * 2 m/s2 * (3.0s)^{2} = 9 m[/tex]

  • So, the work done by the force F can be found as follows:
  • W = F * d = 580 N* 9 m = 5,220 J
  • (An identical outcome could have been found applying the work-energy theorem).

C)

  • The instantanous power can be calculated as follows:

        [tex]P =\frac{W}{\Delta t}[/tex]

  • Replacing W= 5,220 J and t = 3.0 s, we have:

       [tex]P = \frac{W}{\Delta t} = \frac{5220 J}{3.0s} = 1,740 W[/tex]

  • The instantaneous power, at the end of the 3.0 s, is 1,740 W.

D)

  • We can find the instantaneous power at t=1.5 s, using the same equation as above:

       [tex]P =\frac{W}{\Delta t}[/tex]

  • Now, we need first to know the work done during the first 1.5s.
  • We need to find the displacement at the end of the 1.5s time:

        [tex]x = \frac{1}{2} * a * t^{2} = \frac{1}{2} * 2 m/s2 * (1.5s)^{2} = 2.25 m[/tex]

  • We can find now the work done at the end of the first 1.5s, as follows:
  • W = F * d = 580 N * 2.25 m = 1,305J
  • The instantaneous power can be found as follows:

        [tex]P = \frac{W}{\Delta t} = \frac{1,305 J}{1.5.0s} = 870 W[/tex]

  • The instantaneous power, at the end of the first 1.5s, is 870 W.

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