Physics, 25.02.2020 17:58 Leffew

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?

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Answer from: o9623199

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

Explanation:

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

F = m*a (1)

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

$a =\frac{v_{}-v_{0}}{t-t_{0}}$

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

$a =\frac{6.0m/s}{3.0s} = 2 m/s2$

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

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

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

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:

$x = \frac{1}{2} * a * t^{2}$

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

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

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).

The instantanous power can be calculated as follows:

$P =\frac{W}{\Delta t}$

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

$P = \frac{W}{\Delta t} = \frac{5220 J}{3.0s} = 1,740 W$

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

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

$P =\frac{W}{\Delta t}$

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:

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

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 = 1,305JThe instantaneous power can be found as follows:

$P = \frac{W}{\Delta t} = \frac{1,305 J}{1.5.0s} = 870 W$

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

Answer from: Quest

c. magnitude and direction

Answer from: Quest

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.

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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 i...

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