A 7 kg block slides on a horizontal frictionless surface with a speed of 1.2 m/s. It is brought momentarily to rest when it hits a bumper that cmopresses a spring. The acceleration of gravity is 9.8m/s^2.
How much is the spring compressed if its force constant is 3kN/m?
Answer in units of m.
To find the spring compression, we need to use the principle of conservation of energy.
When the block hits the bumper, its initial kinetic energy is converted into the potential energy stored in the compressed spring. Therefore, we can equate these two forms of energy:
(1/2)mv^2 = (1/2)kx^2
Where:
m = mass of the block = 7 kg
v = initial velocity of the block = 1.2 m/s
k = force constant of the spring = 3 kN/m = 3000 N/m (since 1 kN is equal to 1000 N)
x = compression distance of the spring (what we need to find)
We can rearrange the equation to solve for x:
x^2 = (mv^2) / k
x^2 = (7 kg * (1.2 m/s)^2) / (3000 N/m)
x^2 = (7 kg * 1.44 m^2/s^2) / 3000 N/m
x^2 = 0.0336 m^2/s^2 / (3000 N/m)
x^2 = 1.12e-5 m^2
Taking the square root of both sides, we get:
x = √(1.12e-5 m^2)
x ≈ 0.00335 m
Therefore, the spring compression is approximately 0.00335 meters (or 3.35 mm).