Initially sliding with a speed of 2.1 m/s, a 2.1 kg block collides with a spring and compresses it 0.35 m before coming to rest. What is the force constant of the spring?
InitialKE= 1/2 m v^2
1/2 kx^2= KE=1/2 mv^2
k= mv^2/x^2
To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law is given by the equation:
F = -k * x
Where:
F = force exerted by the spring (in Newtons)
k = force constant (also known as spring constant) of the spring (in Newtons per meter)
x = displacement of the spring from its equilibrium position (in meters)
In this case, the block compresses the spring by 0.35 m before coming to rest. Since the spring is compressed, the displacement is negative. Therefore, we can rewrite the equation as:
F = k * (-x)
The block comes to rest when the net force on it is equal to zero. So, we can equate the force exerted by the spring to zero:
0 = k * (-x)
Now, we can rearrange the equation to solve for the force constant (k):
k = 0 / (-x)
Since 0 divided by any number is 0, we conclude that the force constant (k) of the spring is 0.
Therefore, the force constant of the spring is 0 N/m.