Question

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 AWhat is the magnitude of the force exerted by the dogs on the sled?Part BWhat is the work done by the dogs on the sled in the 3.0 s?Part CWhat is the instantaneous power of the dogs at the end of the 3.0 s?Part DWhat is their instantaneous power at 1.5 s?

Answer 1

(a)
`a=2m/sec^2`

(b) 5220 j

(c) 1740 watt

(d) 3446.66 watt

Step-by-step explanation:

We have given mass m = 290 kg

Initial velocity u = 0 m/sec

Final velocity v = 6 m/sec

Time t = 3 sec

From first equation of motion

v = u+at

So
`a=(v-u)/(t)=(6-0)/(3)=2m/sec^2`

(a) We know that force is given by

F = ma

So force will be
`F=290* 2=580N`

(b) From second equation of motion we know that

`s=ut+(1)/(2)at^2=0* 3+(1)/(2)* 2* 3^2=9m`

We know that work done is given by

W = F s = 580×9 =5220 j

(c) Time is given as t = 3 sec

We know that power is given as

`P=(W)/(t)=(5220)/(3)=1740Watt`

(d) Time t = 1.5 sec

So
`P=(W)/(t)=(5220)/(1.5)=3466.66Watt`