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/s?, how
much resistive force (force of friction) is acting on them?

NEED HELP LIKE RIGHT NOW PLz

Answer 1

f = 485.62 N

Step-by-step explanation:

Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:

Unbalanced Force = Horizontal Component Applied Force - Frictional Force

Unbalanced Force = Fx - f

But, from Newtons Second Law of Motion:

Unbalanced Force = ma

comparing the equations:

ma = Fx - f

f = F Cos θ - ma

where,

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°

Therefore,

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

f = 513.55 N - 27.93 N

f = 485.62 N