Question

a 2000 kg car is being pulled up a 37 deg incline at a constant speed by a cable parallel to the incline. if the friction force is 800N WHAT IS THE WORK done by the cable on the car

Answer 1

$$W=mgh+F_fs\\W=(2000)(9.8)(8)\sin{37}+800(8)\\=10^5\ J=100\ kJ$$