Two blocks of mass m 1 = 1.60 kg and m 2 = 4.90 kg are connected by a massless c

Question

Two blocks of mass

m1 = 1.60 kg

and

m2 = 4.90 kg

are connected by a massless cord passing over a frictionless pulley as shown in the figure below, with

= 49.0°.

The coefficient of kinetic friction between block 1 and the horizontal surface is 0.400, and the coefficient of kinetic friction between block 2 and the inclined surface is 0.330.

(a) Draw a free-body diagram for the m1 block.

Draw a free-body diagram for the m2 block.

(b) What is the acceleration of the system when it is released from rest?
m/s2

(c) What is the tension in the cord connecting the blocks?
N

Explanation / Answer

a] i)second figure (top-right one) is correct,

friction Fu = um1g = 0.4m1g

weight = mg

normal = mg

ii) Fourth figure (bottom right one) is correct.

b] Let acceleration be a and tension be T

T - 0.4m1g = m1a

m2g sin theta - T - 0.33 m2g cos theta = m2a

a = [m2g sin theta - 0.4m1g - 0.33 m2g cos theta]/[m1+m2]

= [4.9*9.8* sin 49 degree - 0.4*1.6*9.8 - 0.33*4.9*9.8* cos 49 degree ] /[1.6+4.9]

= 3.011 m/s^2

c] T = 0.4m1g + m1a

= 0.4*1.6*9.8 + 1.6*3.011

= 11.09 N

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