What is the minimum work needed to push a 900 kg car 420 m up a 17.5° incline?

Question

The effective coefficient of friction is 0.25

Any help appreciated... ASAP please.

Answer 1

Resolving forces perpendicular to the slope:
N = 900gcos17.5
where N is the normal to the surface and g is the acceleration due to gravity, 9.81 m/s2

So, N = 8420.4 Newtons

Resolving forces up the slope:
F = uN + 900gsin17.5
where u is the coeff of friction.

F = (0.25)(8420.4) + 900(9.81)(sin17.5)
F = 4760.0 Newtons

Workdone = Force x distance
Workdone = 4760.0 x 420
Workdone = 1,999,200 J or 1,999.2 kJ