A 0.40kg object is attached to a spring constant 160N/m so that the object is allowed to move on a horizontal frictionless surface. The object is released from rest when the sprin g is compressed 0.15m find:

a) the force on the object and
b) its acceleration at this instant.

F = 160N/m * 0.15m = 24.0 N.

a = F/m = 24.0/0.40 = 60.0 m/s^2.

a) The force on the object can be determined using Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula is given by: F = -kx, where F is the force, k is the spring constant, and x is the displacement.

Given that the spring constant, k, is 160 N/m and the displacement, x, is 0.15 m, we can calculate the force on the object:
F = -kx
F = -(160 N/m)(0.15 m)
F = -24 N

Therefore, the force on the object is 24 N.

b) To find the acceleration of the object, we can use Newton's second law of motion, which states that the force on an object is equal to its mass multiplied by its acceleration. The formula is given by: F = ma, where F is the force, m is the mass, and a is the acceleration.

Given that the force, F, is -24N and the mass, m, is 0.40 kg, we can calculate the acceleration:
-24N = (0.40 kg) * a
a = -24N / 0.40 kg
a = -60 m/s^2

Therefore, the acceleration of the object is -60 m/s^2 (negative sign indicates that it is in the opposite direction of the force applied).

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

Hooke's Law equation is given by:

F = -kx

Where:
F = force exerted by the spring
k = spring constant
x = displacement from the equilibrium position

a) The force on the object:
In this case, the object is compressed, so the displacement "x" is negative. Plugging in the given values, we have:

F = -(160 N/m)(-0.15 m)
F = 24 N

Therefore, the force exerted by the spring on the object is 24 N.

b) The acceleration at this instant:
The net force acting on the object is the force exerted by the spring. Since the object is allowed to move on a frictionless surface, there are no other forces acting on it. According to Newton's second law of motion, the net force is equal to the mass of the object multiplied by its acceleration:

F = ma

Rearranging the equation to solve for acceleration:

a = F/m

Plugging in the given values, we have:

a = 24 N / 0.40 kg
a = 60 m/s²

Therefore, the acceleration of the object at this instant is 60 m/s².

A 0.40-kg object is attached to a spring with force constant 160 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 0.15 m. Find a) the force on the object (b) its acceleration at that instant

a. F = 160N/m * 0.15m = 22.5 N.

b. a = F/m = 22.5/0.40 = 56.25 m/s^2.