A 0.49 kg object is attached to a spring with a spring constant 167 N/m so that the object is allowed to move on a horizontal frictionless surface. The object is released from rest when

the spring is compressed 0.12 m. Find the force on the object.

F=kx=167*.12

but F= ma, solve for a.

b) What is the acceleration at this instant?

To find the force on the object, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

The equation for Hooke's Law is:

F = -kx

Where:
F is the force exerted by the spring
k is the spring constant
x is the displacement of the spring

In this case, the object is released from rest when the spring is compressed 0.12 m. Since the object is released, it means it will move in the opposite direction, causing the spring to elongate.

Therefore, the displacement (x) in Hooke's Law equation is -0.12 m (negative sign indicates elongation).

Using the given spring constant (k = 167 N/m) and displacement (x = -0.12 m), we can calculate the force exerted by the spring on the object.

F = -kx
F = -(167 N/m)(-0.12 m)
F = 20.04 N

So, the force on the object is 20.04 N.