Physics question?

Question

A car weighing 8600 N moves along a level highway witha speed of 83 km/h. The power of the engine at this speed is 67kW. The car encounters a hill inclined at an angle of 5.9 degrees with respect to the horizontal. If there is no change in the power of the engine, and no change in the resistive forces acting on the car, what is the new speed of the car on the hill? Answer in units of km/h, and show steps because i want to know how to do this for future use.

Answer 1

If the highway is level the car engine power is wasted against friction only to keep the constant speed of the car. When there’s some inclination of b=5.9 the power N=67000W is shared between speed and gravity. Let v1 be = 83000/3600 = 23.05555m/s
Hence the work of the engine on the level highway for 1s:
A=N*1s=P*k*(v1*1s), v1*1s – path of the car, k – friction resistance, P=8600N.
Now slope: A=N*1s=A1+A2, (N does not change!), A1 is the work against resistance =P*cos(b)*k*(v2*1s), A2 is the work against gravity =P*(v2*1s)*sin(b), where v2*1s – path of the car up slope, (v2*1s)*sin(b) – height of the slope.
Equation: A=N*1s=P*k*(v1*1s)= P*cos(b)*k*(v2*1s)+ P*(v2*1s)*sin(b), or
P*k*v1= P*cos(b)*k*v2+ P*v2*sin(b), or v1*(P*k*v1)= v1*[P*cos(b)*k*v2+ P*v2*sin(b)], or (as P*k*v1=N)
v1*N=cos(b)*v2*N+v1*P*v2*sin(b),
hence
v2=v1*N/[N*cos(b)+v1*P*sin(b)]=
=17.75(m/s)*3600/1000=63.9km/h,
Never believe man, check it!!!