Question Detail
Question
A push of 80 N moves a 5 kg block up a 30 degree inclined plane. The coefficient of kinetic friction is 0.25, and the length of the plane is 20 m. Compute the work done by the resultant force acting on the block.
Correct Answer
Option A is the correct answer.
Detailed Explanation
Weight component down incline = mg sin θ = 5×9.8×0.5 = 24.5 N. Normal force = mg cos θ = 5×9.8×0.866 = 42.434 N. Friction = μN = 0.25×42.434 = 10.6085 N. Net force up incline = applied force - (weight component + friction) = 80 - (24.5 + 10.6085) = 80 - 35.1085 = 44.8915 N. Work = F_net × d = 44.8915 × 20 = 897.83 J. Not matching. If g=10: mg sin30=25 N, mg cos30=43.3 N, friction=10.825 N, net=80-35.825=44.175 N, work=883.5 J. Option D is 883.6 J.
Hint
Work = net force × distance.