Question

Please Help

A car of mass 400 kais sped up from 10m/s to 30m/s in a time of 15 seconds

What is the starting KE?
what is the ending KE?
what is the work done to speed up the car?
what is the power of watts?​

Answer 1

Step-by-step explanation:

The initial kinetic energy
`KE_0` is

`KE_0 = (1)/(2)mv_0^2 = (1)/(2)(400\:\text{kg})(10\:\text{m/s})^2`

`\:\:\:\:\:\;\:= 2×10^4\:\text{J} = 20\:\text{kJ}`

The final kinetic energy
`KE` is

`KE = (1)/(2)mv^2 = (1)/(2)(400\:\text{kg})(30\:\text{m/s})^2`

`\:\:\:\:\:\:\:= 1.8×10^5\:\text{J} = 180\:\text{kJ}`

The work done W on the car is

`W = \Delta{KE} = KE - KE_0`

`\:\:\:\:\:\:\:= 180\:\text{kJ} - 20\:\text{kJ} = 1.6×10^5\:\text{J}`

The power expended P is

`P = (W)/(t) = \frac{1.6×10^5\:\text{J}}{15\:\text{s}} = 10667\:\text{Watts}`

`\:\:\:\:\:= 10.7\:\text{kW}`