Initially sliding with a speed of 2.3 m/s, a 2.0 kg block collides with a spring and compresses it 0.41 m before coming to rest. What is the force constant of the spring?
1/2 kx^2=1/2 m v^2
solve for k
1/2(2)(2.3)^2=5.29
To find the force constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.
According to Hooke's Law, the force exerted by a spring can be calculated using the formula:
F = -kx
Where:
F is the force exerted by the spring
k is the force constant of the spring
x is the displacement of the spring from its equilibrium position
In this case, the block compresses the spring by 0.41 m, which is the displacement (x) in the formula. The force exerted by the spring is opposing the motion of the block.
Since the block comes to rest, the force exerted by the spring is equal in magnitude and opposite in direction to the initial kinetic energy of the block.
To calculate the force exerted by the spring, we can use Newton's second law:
F = ma
Where:
F is the force exerted by the spring
m is the mass of the block
a is the acceleration of the block
Initially, the block is sliding, so its acceleration is given by:
a = (v^2 - u^2) / (2x)
Where:
v is the final velocity of the block (0 m/s, as it comes to rest)
u is the initial velocity of the block (2.3 m/s)
x is the displacement of the spring (0.41 m)
Plugging in the values:
a = (0^2 - 2.3^2) / (2 * 0.41)
a = -13.9 m/s^2
Using Newton's second law, we can calculate the force exerted by the spring:
F = ma
F = 2.0 kg * -13.9 m/s^2
F = -27.8 N
Since the force exerted by the spring is negative, it indicates that the force is directed opposite to the compression direction.
Now, substituting the force and displacement values into Hooke's Law, we can calculate the force constant of the spring:
-27.8 N = -k * 0.41 m
Dividing both sides of the equation by 0.41 m, we get:
k = 67.80 N/m
Therefore, the force constant of the spring is 67.80 N/m.