A block with mass m = 0.250 kg is attached to one end of an ideal spring and mov

Question

A block with mass m = 0.250 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 = -13.0 m/s2 and its velocity is vx = +4.00 m/s.

What is the spring's force constant k?

The correct answer is 13.5 N/m.

What is the amplitude of the motion?

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

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

Explanation / Answer

Here ,

for the block ,

angular frequency , w = sqrt(k/m)

w = sqrt(13.5/0.25)

w = 7.35 rad/s

let the ampitude is A

Using conservation of energy

0.5 * k * A^2 = 0.5 * m * v^2 + 0.5 * k * x^2

13.5 * A^2 = 13.5 * 0.24^2 + 0.25 * 4^2

solving for A

A = 0.595 m

the amplitude of motion is 0.595 m

---------------------------

maximum speed = A * w

maximum speed = 0.595 * 7.35

maximum speed = 4.37 m/s

---------------------

maximum magnitude of the block's acceleration = A * w^2

maximum magnitude of the block's acceleration = 0.595 * 7.35^2

maximum magnitude of the block's acceleration = 32.13 m/s^2

the maximum magnitude of the block's acceleration is 32.13 m/s^2

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