A block of mass M=6 kg and initial velocity v=0.8m/s slides on a frictionless horizontal surface and collides with a relaxed spring of unknown spring constant. The other end of the spring is attached to a wall. If the maximum compression of the spring is 0.2 m, what is the spring constant?

Equate kinetic and potential (spring) energies:

KE=(1/2)mv²
PE=(1/2)ks²
Solve for k
k=mv²/s² N/m
(verify that quantities are in consistent units).

I tried

F=-kx so
mg=-kx
(6 kg)(9.8m/s^2)=-k(0.2m)
k= 294 N/m

But the correct answer is supposed to be 96 Nm.

Oh I see now. Thanks!

You're welcome!

To find the spring constant of the relaxed spring, we can use the principle of conservation of mechanical energy.

1. The first step is to determine the potential energy stored in the spring when it is at maximum compression.

When the spring is at maximum compression, it comes to a temporary stop for an instant before bouncing back. At this point, all the initial kinetic energy of the sliding block is converted into potential energy stored in the spring.

The potential energy stored in a spring can be calculated using the formula:

Potential Energy (Elastic) = (1/2) * k * x^2

Where:
- k is the spring constant (what we want to find)
- x is the displacement or compression of the spring (0.2 m in this case)

For maximum compression, the block momentarily comes to rest, so it has no kinetic energy. Hence, all its energy is stored in the spring as potential energy.

Potential Energy (Elastic) = Initial Kinetic Energy of the block

2. Calculate the initial kinetic energy of the block.

The initial kinetic energy of the block is given by the formula:

Kinetic Energy = (1/2) * M * v^2

Where:
- M is the mass of the block (6 kg in this case)
- v is the initial velocity of the block (0.8 m/s in this case)

3. Equate the potential energy stored in the spring to the initial kinetic energy of the block and solve for the spring constant.

Equate the expressions for potential energy and initial kinetic energy:

(1/2) * k * x^2 = (1/2) * M * v^2

Since we are solving for the spring constant, rearrange the equation:

k = (M * v^2) / x^2

Now, substitute the known values:

k = (6 kg * (0.8 m/s)^2) / (0.2 m)^2

Simplifying:

k = (6 kg * 0.64 m^2/s^2) / 0.04 m^2

k = 96 N/m

So, the spring constant of the relaxed spring is 96 N/m.