A block of mass m = 7.0 kg is dropped from a height H = 46.0 cm onto a spring of spring constant k = 2360 N/m (see the figure). Find the maximum distance the spring is compressed.

mgh=kx²/2

x=sqrt(2mgh/k)

0.164 To find the maximum distance the spring is compressed, we need to apply the principles of conservation of mechanical energy.

Step 1: Calculate the gravitational potential energy of the block.
The potential energy of an object at a certain height is given by:

PE = m * g * h

Here, m is the mass of the block, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from which the block is dropped.

Given:
m = 7.0 kg (mass of the block)
h = 46.0 cm = 0.46 m (height from which the block is dropped)
g = 9.8 m/s^2 (acceleration due to gravity)

PE = (7.0 kg) * (9.8 m/s^2) * (0.46 m)
PE = 31.732 J (Joules)

Step 2: Find the maximum potential energy stored in the spring.
When the block hits the spring, the potential energy it had due to its height gets converted into potential energy stored in the spring.

The maximum potential energy stored in a spring is given by:

PE = (1/2) * k * x^2

Here, k is the spring constant and x is the maximum distance the spring is compressed.

Given:
k = 2360 N/m (spring constant)

31.732 J = (1/2) * (2360 N/m) * x^2

Simplifying this equation will give us the value of x, which is the maximum distance the spring is compressed.

Step 3: Solve for x.
To solve for x, we rearrange the equation:

x^2 = (2 * PE) / k

x^2 = (2 * 31.732 J) / 2360 N/m

x^2 = 0.02701 m^2

Taking the square root of both sides:

x = 0.1645 m

Therefore, the maximum distance the spring is compressed is approximately 0.1645 meters.