For the total work, E = -(1/2)*m*v0^2 = -3.8553*10^8 J
the engine work, W = -F0*s = -3.6606*10^8 J
so delta E = E - W = -1.947*10^7 J.
For the integration,
F/m = -dv/dt, =>
v(t) = -∫(F/m)dt
= -∫[-(F0+F1)/m + F1*t/(m*ts)]dt
= -(1/m)*[-(F0+F1)*t + (1/2)*F1*(t^2)/ts]
so...
Hi, guys. Is the magnitude of friction work equal to heat absorbed by the wheels?
I calculated the total work and the engine work, then the difference delta E between them should be friction work. I also calculated the friction work Ef alone with integration. Then I can see that delta E = Ef...