An 6.82 kg block drops straight down from a height of 1.31 m, striking a platform spring having a force constant of 1.16 103 N/m. Find the maximum compression of the spring

You must mean k = 1.16*10^3 N/m

The decrease in gravitational P.E. equals the stored spring P.E when it stops (at maximum compression, X).

M g (1.31 + X) = (1/2) k X^2

The X is on the left because gravity does additional work during compression. A quadratic equation solution will be needed. Your teacher may expect you to neglect the X term and get a simpler (but approximate) answer.

To find the maximum compression of the spring, we can use the conservation of mechanical energy. The potential energy lost by the block as it falls is equal to the elastic potential energy stored in the compressed spring.

The potential energy lost by the block can be calculated using the formula:

Potential energy lost = m * g * h

Where:
m = mass of the block = 6.82 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height from which the block is dropped = 1.31 m

Let's calculate the potential energy lost:

Potential energy lost = 6.82 kg * 9.8 m/s^2 * 1.31 m

Next, we can equate this potential energy lost to the elastic potential energy stored in the compressed spring. The elastic potential energy stored in a spring can be calculated using the formula:

Elastic potential energy = (1/2) * k * x^2

Where:
k = force constant of the spring = 1.16 * 10^3 N/m
x = maximum compression of the spring (what we need to find)

Setting the potential energy lost equal to the elastic potential energy, we have:

Potential energy lost = Elastic potential energy
6.82 kg * 9.8 m/s^2 * 1.31 m = (1/2) * 1.16 * 10^3 N/m * x^2

Now, let's solve for x:

6.82 kg * 9.8 m/s^2 * 1.31 m = (1/2) * 1.16 * 10^3 N/m * x^2

Simplifying the equation, we find:

x^2 = (6.82 kg * 9.8 m/s^2 * 1.31 m) / (0.5 * 1.16 * 10^3 N/m)

Finally, we can solve for x by taking the square root of both sides of the equation:

x = √ [(6.82 kg * 9.8 m/s^2 * 1.31 m) / (0.5 * 1.16 * 10^3 N/m)]

Evaluating this expression will give us the maximum compression of the spring.