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A block, whose mass is 0.500 kg, is attached to a spring with a force constant of 126 N/m. The block rests upon a frictionless, horizontal surface. A block labeled m is attached to the right end of a horizontal spring, and the left end of the spring is attached to a wall. The spring is stretched horizontally such that the block is displaced by a distance A to the right of its equilibrium position. The block is pulled to the right a distance A = 0.120 m from its equilibrium position (the vertical dashed line) and held motionless. The block is then released from rest. (a) At the instant of release, what is the magnitude of the spring force (in N) acting upon the block? N (b) At that very instant, what is the magnitude of the block's acceleration (in m/s2)? m/s2 (c) In what direction does the acceleration vector point at the instant of release?

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hamzaahmeds

Answer:

(a) F = 15.12 N

(b) a = 30.24 m/s²

(c) To Left

Explanation:

(a)

The magnitude of the spring force is given by Hooke's Law as follows:

F = kx

where,

F = Spring Force = ?

k = Spring Constant = 126 N/m

x = Displacement = A = 0.12 m

Therefore,

F = (126 N/m)(0.12 m)

F = 15.12 N

(b)

The magnitude of acceleration can be found by comparing the spring force with the unbalanced force formula of Newton's Second Law:

F = ma

where,

F = Spring Force = 15.12 N

m = mass of block = 0.5 kg

a = magnitude of acceleration = ?

15.12 N = 0.5 kg (a)

a = 15.12 N/0.5 kg

a = 30.24 m/s²

(c)

Since, the acceleration is always directed towards mean (equilibrium) position in periodic motion. Therefore, the direction of the acceleration at the time of release will be to left.

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