Question

When a car of mass 1079 kg accelerates from 3.00 m/s to some finals speed, 2.00x10^5 J of work are done. Find the final speed

Answer 1

Explanation:

The work-energy theorem says that the change in kinetic energy of a system is equal to the work done.

`(1)/(2)mv^2_f-(1)/(2)mv^2_i=\Delta W^{}`
`(1)/(2)m(v^2_f-v^2_i)=\Delta W^{}`

Now in our case

Therefore, the above equation gives

`(1)/(2)(1079)(v^2_f-(3.00)^2^{}_{})=2.00*10^5J`

Now we need to solve for v_f .

Mutlipying both sides by 2 gives

`(1079)(v^2_f-(3.00)^2_{})=2.00*10^5*2`

dividing both sides by 1079 gives

`(v^2_f-(3.00)^2_{})=(2.00*10^5J)/(1079)*2`

Finally, adding 3.00^2 to both sides gives

`v^2_f=(2.00*10^5J)/(1079)*2+3.00^2`

Finally, simplifying the right-hand side gives

`v^2_f=379.71`

taking the square root of both sides gives

`\boxed{v_f=19.49m/s}`

Hence, the final speed of the car is 19.49 m/s.