) A 4.8 kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t = 0 s, the block has a displacement of -0.90 m a velocity of -0.80 m/s and an acceleration of +2.9 m/s^2. The force constant of the spring is closest to:
To determine the force constant of the spring, we can make use of 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.
The equation representing Hooke's Law is:
F = kx
Where:
F = force exerted by the spring (in newtons, N)
k = force constant of the spring (in newtons per meter, N/m)
x = displacement of the spring from its equilibrium position (in meters, m)
In this case, we are given the mass of the block (4.8 kg), its displacement (-0.90 m), velocity (-0.80 m/s), and acceleration (+2.9 m/s^2) at time t = 0 s.
First, let's calculate the net force acting on the block at t = 0 s by using Newton's second law:
F_net = ma
Where:
F_net = net force acting on the block (in newtons, N)
m = mass of the block (in kilograms, kg)
a = acceleration of the block (in meters per second squared, m/s^2)
Plugging in the given values, we have:
F_net = (4.8 kg) * (+2.9 m/s^2)
F_net = 13.92 N
Since the block is attached to the spring, the net force acting on the block is equal to the force exerted by the spring:
F_net = F = kx
We can rearrange this equation to solve for the force constant, k:
k = F_net / x
Plugging in the values, we get:
k = 13.92 N / (-0.90 m)
k = -15.47 N/m
Therefore, the force constant of the spring is approximately -15.47 N/m.