An ice block of mass M=14.0 kg, is being pulled up a ramp with a force F=170 N m

Question

An ice block of mass M=14.0 kg, is being pulled up a ramp with a force F=170 N making an angle ?=32.0° with the ramp. The ramp makes the same angle ? with the horizontal. The ice block has an initial speed of 1.60 m/s and is pulled a distance of 8.70 m. Assume that the coefficient of kinetic friction between the ramp and the block is 0.3. (Use 10.0 N/kg for g)

a)0 how many forces acting on the ice block ?

b) How many forces are parallel to the ramp?

c) What is in degrees (as a positive number measured in the counterclockwise direction, relative to the positive horizontal x-direction -note that it is the horizontal x-direction not the direction parallel to the ramp) the direction of the normal force on the ice block?

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

e) What is the magnitude of the net force acting on the ice block?

Explanation / Answer

Mass M=14.0 kg,
Force F=170 N
uk = 0.3
=32.0°

(A)
Forces acting on the ice block -
External Force F is acting at an angle .
Horizontal component of External Force parallel to the incline upwards = F*cos()

Vertical Component = F*sin()
Normal Force Perpendicular to the incline, Fn = (Mg*cos() - F*sin())

Friction Force , parallel to the incline in the downward direction, Fr = uk*Fn

Weight of Block along the incline = Mg*sin()

Total Forces, = 4.
(b)
Forces are parallel to the ramp = 3
Friction Force.
Parallel component of External Force F.
Force due to weight of the block along the ramp.

(c)
Direction of the normal force on the ice block = 90 + = 90 + 32 = 1220.

(d)
Magnitude of Normal Force, Fn = (Mg*cos() - F*sin())
Fn = 14*10*cos(32) - 170*sin(32)
Fn = 28.64 N

(e)
Magnitude of Net Force = F*cos(32) - Fr - Mg*sin(32)

Fnet = F*cos(32) - uk*Fn - Mg*sin(32)
Fnet = 170*cos(32) - 0.3*28.64 - 14*10*sin(32)
Fnet = 61.38 N

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