Agirl pushes a 10.0kg sled at a constant speed by applying a force of 75n at an angle of 30 degrees with respect to the horizontal. the sled is pushed over a distance of 15 meters. how much work is done by the force of friction?

Answer

given,

mass of sled = 10 Kg

Force applied by girl = ?

angle of inclination = 30°

distance = 15 m

Sled is moving with constant speed hence acceleration will be equal to zero.

now,

F - mg sin θ - f = m a

F - mg sin θ - f = 0

F - mg sin θ = f

where f is the frictional force acting opposite to the applied force

mg sin θ is the component of the weight which is opposing the motion of the sled

now,

f = F - mg sin θ

f = 75 - 10 x 9.8 x sin 30°

f = 26 N

now,

Work done = f. s cos ∅

where ∅ is the angle between frictional  force and displacement which is equal to 180°

W= 26 x 15 x cos 180°

W = -390 J

work done by frictional force is equal to W = -390 J

i think the answer is b