Question

A 0.37 kg object is attached to a spring with a spring constant 175 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 0.19 m. Find the force on the object when it is released. Answer in units of N. What is the acceleration at this instant? Answer in units of m/s^2.​

Answer 1

The force exerted on a 0.37 kg object when released from a compressed spring with a constant of 175 N/m is 33.25 N. The acceleration at this instant is 89.86 m/s².

Step-by-step explanation:

We are given a 0.37 kg object attached to a spring with a spring constant of 175 N/m, compressed by 0.19 m. The force exerted by the spring when it is released can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position:

F = -kx

Where:

• F is the force in newtons (N),
• k is the spring constant in newtons per meter (N/m), and
• x is the displacement from the equilibrium position in meters (m).

Plugging in the values we get:

F = -(175 N/m) * (0.19 m)

F = -33.25 N

Since the negative sign indicates the direction of the force is opposite to the direction of displacement, we can say that the magnitude of the force is 33.25 N. To find the acceleration at this instant, we can use Newton's second law of motion:

a = F/m

a = 33.25 N / 0.37 kg

a = 89.86 m/s²

The acceleration of the object when it is released is 89.86 m/s².

Answer 2

`F=33.25\ N`

`a=89.8649\ m.s^(-2)`

Step-by-step explanation:

Given:

• mass of the object,
  `m=0.37\ kg`

• spring constant,
  `k =175\ N.m^(-1)`

• compression in the spring,
  `\Delta x=0.19\ m`

Now the force on the spring on releasing the compression:

`F=k. \Delta x`

`F=175* 0.19`

`F=33.25\ N`

Now the acceleration due to this force:

`a=(F)/(m)`

`a=(33.25)/(0.37)`

`a=89.8649\ m.s^(-2)`