Two blocks are connected by a light cord passing over a small, frictionless pulley as shown in the figure. Block A has mass 5.00 kg and block B has mass 4.00 kg. If block B descents at a constant acceleration of 2.00 m/s^2 when set in motion, what is the coefficient of kinetic friction between block A and the table. 0.23 1.02 0.12 0.21 0.43 O O O

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Two blocks are connected by a light cord passing over a small, frictionless pulley as shown in
the figure. Block A has mass 5.00 kg and block B has mass 4.00 kg. If block B descents at a
constant acceleration of 2.00 m/s^2 when set in motion, what is the coefficient of kinetic
friction between block A and the table.
0.23
1.02
0.12
0.21
0.43
O O O
Transcribed Image Text:Two blocks are connected by a light cord passing over a small, frictionless pulley as shown in the figure. Block A has mass 5.00 kg and block B has mass 4.00 kg. If block B descents at a constant acceleration of 2.00 m/s^2 when set in motion, what is the coefficient of kinetic friction between block A and the table. 0.23 1.02 0.12 0.21 0.43 O O O
Expert Solution
Step 1

Mechanical Equilibrium: A body is said to be in mechanical equilibrium if it satisfies the following conditions

1. The net force in the x-direction is zero

2. The net force in the y-direction is zero

3.The sum of the moments about any point is zero

Sign conditions: Forces in the positive x-direction are taken as positive and forces in the negative x-direction are taken as negative. 

For the y-axis also same conditions are taken as the x-axis

Clockwise moments are taken as positive and anticlockwise moments are taken as negative. 

Step 2

Given Data: the mass of the block B is mB = 4 kg 

The mass of block A is mA = 5 kg 

Let T be the tension in the cord 

Acceleration a = 2 m/s2

Drawing the free body diagram of the block B

Mechanical Engineering homework question answer, step 2, image 1

From the diagram Calculating the net force in the y-direction T - mBg = - mBaT - 4*9.81  = -4*2T = 31.24 N 

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