
Concept explainers
(a)
The acceleration of each blocks and tension in the cord.
(a)

Answer to Problem 122P
The magnitude of acceleration of each blocks and tension in the cord is respectively
Explanation of Solution
Case 1 (mass
Write the expression for the equilibrium condition in y-direction (Vertical).
Here, N is the normal force on the mass
Write the expression for the equilibrium condition in x-direction (Horizontal).
Here,
Rewrite the above equation in terms of
Case 2 (mass
Write the expression for the equilibrium condition in y-direction (Vertical).
Here,
Write the above expression in terms of
Since both acceleration is equal in magnitude and opposite in direction.
Write the expression for the relation between acceleration of two blocks.
Substitute (III) and (V) in the above equation to calculate T.
Substitute (VI) in (v) to calculate the acceleration
Substitute the above expression in (VI) to calculate
Conclusion:
Therefore, the magnitude of acceleration of each blocks and tension in the cord is respectively
(b)
The magnitude of acceleration and tension for the cases
(b)

Answer to Problem 122P
The magnitude of acceleration and tension for the cases
Explanation of Solution
Consider the equation (VII) and (VIII).
Case 1
Since
Thus, the tension in the cord is
Since
Thus, the acceleration of the block is
Case 2
Since
Thus, the tension in the cord is
Since
Thus, the acceleration of the block is
Case 3
Substitute
Conclusion:
Therefore, the magnitude of acceleration and tension for the cases
(c)
The magnitude of
(c)

Answer to Problem 122P
The magnitude of
Explanation of Solution
Consider the equation (VII) and (VIII).
In the case of constant velocity acceleration should be zero.
Equate (VIII) to zero to calculate
Substitute the above equation for
Conclusion:
Therefore, the magnitude of
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Chapter 4 Solutions
Physics
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