High School

A 0.40 kg object is attached to a spring with a force constant of 157 N/m, allowing the object to move on a horizontal frictionless surface. The object is released from rest when the spring is compressed 0.17 m.

(a) Find the force on the object.
(b) Find its acceleration at that instant.

Answer :

The force on the object is -26.69 N with the negative sign indicating that the force is directed towards the equilibrium position. Its acceleration at that instant is equal to 66.725 m/s².

(a) Find the force on the object

Hooke's Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position and is given by:

F = -kx

where F is the force exerted by the spring, k is the force constant (spring constant), and x is the displacement from the equilibrium position.

Given:

k = 157 N/m

x = 0.17 m

Plugging in the values, we get:

F = -157 N/m × 0.17 m = -26.69 N

The negative sign indicates that the force is directed towards the equilibrium position.

(b) Find its acceleration at that instant

Using Newton's Second Law of Motion, we have:

F = ma

where F is the force, m is the mass, and a is the acceleration.

Given:

m = 0.40 kg

F = 26.69 N (we use the magnitude)

Plugging in the values, we get:

26.69 N = 0.40 kg × a

a = 26.69 N / 0.40 kg = 66.725 m/s²

Therefore, the acceleration of the object at that instant is 66.725 m/s².

Final answer:

The force on the object when the spring is compressed is -26.69 N, and the acceleration at that instant is -66.725 m/s². These calculations involve the principles of Hooke's Law and Newton's second law.

Explanation:

This problem involves the principles of mechanics and Hooke's Law, related to the behavior of springs. When a spring is compressed or extended, it exerts a force to restore itself to its normal length. This force, according to Hooke's Law, is F = -kx, where k is the spring constant and x is the displacement.

(a) The force exerted by the spring when it is compressed by 0.17 m would be calculated by F = -kx = -157 N/m * 0.17 m = -26.69 N. This is negative, indicating that the direction of the force is opposite to the direction of displacement, aiming to restore the spring to its normal length.

(b) The acceleration of the object at that instant can be determined using Newton's second law, F = ma. Rearranging, the acceleration a = F/m = -26.69 N / 0.40 kg = -66.725 m/s². The negative sign indicates the acceleration is in the opposite direction of the initial displacement.

Learn more about Hooke's Law here:

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