Question

A 49 kg person is being dragged in their sleeping bag to the lake by a 593 N

force at an angle of 30°. If the person accelerates at a rate of 0.57 m/s2, how
much resistive forte (force of friction) is acting on them?

Answer 1

485.62 N

Step-by-step explanation:

To obtain the magnitude of unbalanced forces acting on the body;

Unbalanced Force = Horizontal Component of Applied Force - Frictional Force

Frictional Force = Horizontal Component of Applied Force - Unbalanced Force

f = frictional force = ?

F = Applied force = 593 N

m = mass of person = 49 kg

a = acceleration = 0.57 m/s²

θ = Angle with horizontal = 30°

Hence;

Horizontal Component of Applied Force = (593 N)(Cos 30°)

Unbalanced Force = (49 kg)(0.57 m/s²)

f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)

f = 513.55 N - 27.93 N

f = 485.62 N