Question

an initially stationary 2kg object accelerates horizontally anduniformly to a speed of 10m/s in 3s
a)in that 3s interval how much work is done on the object by theforce accelerating it?
what is the instantaneous power due to that force b)at the end ofthe interval and c) at the end of the first half of theinterval?

Answer 1

Given that

Mass of the object (m) =2.0kg

Speed(v) =10m/s

Time(t) =3.0s

Then the acceleration is

                 a = v/t

                    =10/3

                    =3.33m/s²

Initial velocity(v₀) = 0m/s

Let us consider it covers a x distance in 3.0s then from theequation of motion we get

                 x =v₀t +(1/2)at²

                    =0 + 15

                    = 15m

a)

The work done is

                      W =F.x

                           =max

                            =(2.0kg)(3.33m/s²)(15m)

                           =100J

b)

At t =3.0s we get the instantaneous power due to that forceis

                      P = F.v

                         = mav

                         = (2kg)(3.33m/s²)(10m/s)

                          =66.6W

c)

At t= 3/2 s we get instantaneous power due to that forceis

                      P =F.v

                          =ma²t

                         =100/3

                         =33.3W

i hope this may helps you