A 6.0kg block is falling toward a spring located 1.80m below. If it has a speed of 4.0m/s at that instant, what is the maximum compression of the spring if the spring constant is is 2.000 x 10^3 N/m?
KE of block + PE of block=max spring energy
1/2 m v^2+mg(1.9)=1/2 *2e3*x^2
put in mass m, v velocity, g and solve for comression x.
To find the maximum compression of the spring, we need to use the principle of conservation of mechanical energy.
The mechanical energy at the top (before the block starts falling) is equal to the sum of the gravitational potential energy and the elastic potential energy when the spring is compressed.
The gravitational potential energy at the top (before falling) can be calculated using the formula:
PE_gravity = m * g * h
Where:
m = mass of the block = 6.0 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height = 1.80 m
PE_gravity = 6.0 kg * 9.8 m/s^2 * 1.80 m
PE_gravity = 105.84 J
At the bottom when the block reaches the spring, all the gravitational potential energy is converted into elastic potential energy stored in the spring. Therefore, we can equate these two energies:
PE_gravity = PE_spring
105.84 J = 1/2 * k * x^2
Where:
k = spring constant = 2.000 x 10^3 N/m
x = maximum compression of the spring (what we want to find)
Rearranging the equation:
1/2 * 2.000 x 10^3 N/m * x^2 = 105.84 J
Simplifying:
x^2 = (2 * 105.84 J) / (2.000 x 10^3 N/m)
x^2 = 0.10584 J / 1.000 N/m
x^2 = 0.10584 m
Taking the square root:
x = sqrt(0.10584 m)
x ≈ 0.3253 m
Therefore, the maximum compression of the spring is approximately 0.3253 meters.