A block with mass m = 0.300 kg is attached to one end of an ideal spring and mov

Question

A block with mass m = 0.300 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The other end of the spring is attached to a wall. When the block is at x = +0.240 m, its acceleration is ax = -14.0 m/s2and its velocity is vx = +4.00 m/s.

A.

What is the spring's force constant k?

Express your answer with the appropriate units.

B.

What is the amplitude of the motion?

Express your answer with the appropriate units.

C.

What is the maximum speed of the block during its motion?

Express your answer with the appropriate units.

D.

What is the maximum magnitude of the block's acceleration during its motion?

Express your answer with the appropriate units.

Explanation / Answer

m = 0.3 kg ; x = +0.24 m ; ax = -14 m/s^2 ; vx = +4 m/s

A)at any point of time

A cos(wt + phi) = 0.24

-A w^2 cos(wt + phi) = -12

A w^2 cos(wt + phi)/A cos(wt + phi) = 12/0.24

w^2 = 50

k/m = 50 => k = 50 x m = 50 x 0.3 = 15 N/m

Hence, k = 15 N/m

B)from conservation of energy:

1/2 k A^2 = 1/2 m v^2 + 1/2 k x^2

A = sqrt [(mv^2 + kx^2)/k]

A = sqrt [(0.3 x 4^2 + 15 x 0.24^2)/15] = 0.614 m

Hence, A = 0.614 m

C)Again using energy conservation

1/2 k A^2 = 1/2 m Vm^2

Vm = sqrt (k A^2/m) = sqrt (15 x 0.614^2/0.3) = 4.34 m/s

Hence, Vm = 4.34 m/s

D)a = w^2 A

am = 50 x 0.614 = 30.7 m/s^2

Hence, am = 30.7 m/s^2

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