Question

If a `2 kg` object is being constantly accelerated from `0m/s` to `15 m/s` over `6s`, how much power is being applied at `t= 1`?

Answer 1

Acceleration = `a=(v-u)/t=(15-0)/6=2.5ms^-2`
velocity after 1s `= v_1= u +axx1=0+2.5xx1=2.5m/s`
Mechanical work (w)=KE gained = `1/2xxmxxv^2=1/2xx2xx(2.5)^2=6.25J`

Power `P = w/t=6.25/1W=6.25W`