An ice block of mass M=12.4 kg, is being pulled up a ramp with a force F=161 N m

Question

An ice block of mass M=12.4 kg, is being pulled up a ramp with a force F=161 N making an angle =26.0° with the ramp. The ramp makes the same angle with the horizontal. The ice block has an initial speed of 1.65 m/s and is pulled a distance of 8.80 m. Assume that the coefficient of kinetic friction between the ramp and the block is 0.2.

a) How many forces are 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

a)

Refer the below figure,

Total 4 forces act on the block.

Fg,F, Fk and Fn

b)

Refer above figure,

Total 3 force are parallel to the ramp,

Fk, Fgsinq, Fx

c) Fn act 90 deg to the horizontal (X axis)

d)

Applying Newton’s second law to vertical

Fn-mgcosq = ma(y)=0

Fn= mgcosq= 12.4*9.8*cos26 = 109.22N

e) Applying Newton’s second law to horizontal

Fnet = Fx-mgsinq-Fk =

Fnet = Fcosq - mgsinq - uk*mgcosq

Fnet = [Fcosq - mgsinq - uk*mgcosq] = [161cos26-12.4*sin26-0.2*12.4*9.8*cos26] = 117.4 N

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