16. (Force and Motion - II) Two blocks of masses m, are suspended by a cord from a pulley which is attached to in front of a wall as shown in figure. A horizontal force of 8.3 N is applied to second block and the coefficients of static and kinetic frictions between the wall and the second block are 0.4 and 0.2. Cord and pulley are massless and pulley is frictionless. 1 kg and m2 = 2 kg %3D If the blocks, which are initially at rest, start moving when they are released; i Draw the free body diagrams for both blocks. m1 m) ii Find the acceleration of the blocks. Answer: ii) a = 4.49 m/s
16. (Force and Motion - II) Two blocks of masses m, are suspended by a cord from a pulley which is attached to in front of a wall as shown in figure. A horizontal force of 8.3 N is applied to second block and the coefficients of static and kinetic frictions between the wall and the second block are 0.4 and 0.2. Cord and pulley are massless and pulley is frictionless. 1 kg and m2 = 2 kg %3D If the blocks, which are initially at rest, start moving when they are released; i Draw the free body diagrams for both blocks. m1 m) ii Find the acceleration of the blocks. Answer: ii) a = 4.49 m/s
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question

Transcribed Image Text:16. (Force and Motion - II) Two blocks of masses m,
are suspended by a cord from a pulley which is attached to in front of
a wall as shown in figure. A horizontal force of 8.3 N is applied to
second block and the coefficients of static and kinetic frictions between
the wall and the second block are 0.4 and 0.2. Cord and pulley are
massless and pulley is frictionless.
1 kg and m2 = 2 kg
%3D
If the blocks, which are initially at rest,
start moving when they are released;
i Draw the free body diagrams for
both blocks.
m1
m)
ii Find the acceleration of the blocks.
Answer: ii) a =
4.49 m/s
Expert Solution

Step 1
Drawing the free body diagram of the bock as shown below –
When the system will be released then there will be tension in the cord. A friction force will be generated between the wall and second block which will act opposite to the direction of motion. A reaction force will act on the wall and second block.
Step 2
Writing the force balance equation as shown below –
Step by step
Solved in 4 steps with 5 images

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