Question

A particle of mass m is lying on smooth horizontal table. A constant force F tangential to the surface is applied on it. Find . (a) average power over a time interval from t=0 to t=t, (b) instantaneous power as function of time t.

Answer 1

Power = (F^2)(t)/(2m)

• Given: Force (F) = F.

• Acceleration = F/m

• Velocity = Ft/m

        Displacement = $\frac{Ft^{2}}{2m}$

        Average Power = $\frac{F^{2}t }{2m}$

• Instantaneous Power:
• P$\int\limits^ {} \, dt$= F.$\int\limits^ {} \, dx$
• P = (F^2)(t)/(2m)

Answer 2

Here  Average power =$F^{2}$*t/m and Instantaneous Power  =$F^{2}$*t/m.

Given:

1.Mass of the particle = m (given)

2.Force (F) = F.

3.Acceleration = F/m    (Since F=m*a)

4.Velocity = Ft/m           (Since a= v/t)

5.Displacement = Ft^2/m         (Since v= displacement/t)

a) Now  Average Power = work done/t

                               = F*displacement/t

                              = F*F*t^2/t*m

                              =$F^{2}$*t/m

b) Instantaneous Power:

1. We know power = work done/t

                               = F*s/t

 

here for an instant s = ds and t= dt

i) So ds/dt = dv

2.Therefore instantaneous power = F*dv

                                                         =F*F*t/m

                                                       =$F^{2}$*t/m