A 11.8 kg crate is pulled by a 148 N force (parallel to the incline) up a rough

Question

A 11.8 kg crate is pulled by a 148 N force (parallel to the incline) up a rough incline. Assume that the incline makes an angle of 24.0° with the horizontal and that the coefficient of kinetic friction is 0.609. The crate has an initial speed of 1.67 m/s. It is pulled a distance of 8.60 m.

a) How many forces are acting on the block?

b) How many forces are parallel to the incline?

c) How many forces are perpendicular to the incline?

d) What is the magnitude of the normal force acting on the block?

***Use 10.0 N/kg for g.***

Please give details.

Explanation / Answer

(a) Force acting on the Block -

External Force , F = 148 N (Parallel to incline)
Frictional Force, Fr = uk*Fn (Parallel to incline, against the motion.)
Normal Force , Fn = mg*cos(24) (Perpendicular to the incline)
Gravitational Force, Fg = mg sin(24) (Along the incline downwards)

Force acting on the Block = 4

(b)
Forces parallel to the incline -
External Force , F = 148 N (Parallel to incline)
Frictional Force, Fr = uk*Fn (Parallel to incline, against the motion.)
Gravitational Force, Fg = mg sin(24) (Along the incline downwards)

Forces parallel to the incline = 3

(c)
Forces Perpendicular to the incline = 1
Normal Force , Fn = mg*cos(24) (Perpendicular to the incline)

(d)
Magnitude of Normal Force , Fn = 11.8 * 10 * cos(24) N
Fn = 107.8 N

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