Answer to Question #84365 in Mechanics | Relativity for Olowu Idowu

Question

A body of mass 10kg and initially at rest is subjected to a force of 20N for 1 second. Calculate the change in kinetic energy for the body during that time

Answer 1

Change in kinetic energy:

∆E_k=E_k2-E_k1=(mϑ^2)/2-(mϑ_0^2)/2=(mϑ^2)/2," "(1)

where (mϑ_0^2)/2=0, because ϑ_0=0.

According to Newton's second law of motion:

F=ma=m (ϑ-ϑ_0)/t=mϑ/t " "(2)

From (2) we can find the final velocity:

ϑ=Ft/m " "(3)

Substituting (3) into (1), we receive:

∆E_k=(mϑ^2)/2=(mF^2 t^2)/(2m^2 )=(F^2 t^2)/2m

∆E_k=(〖20〗^2∙1^2)/(2∙10)=20 (J)