Answer: 2.53 m/s
Explanation: The first step in solving this problem is to use Hooke's law to find the force exerted by the spring on the mass when it is stretched by 4 cm. Hooke's law states that the force exerted by a spring is proportional to the amount it is stretched or compressed. The formula for Hooke's law is:
F = -kx
where F is the force exerted by the spring, k is the force constant of the spring, and x is the amount the spring is stretched or compressed from its equilibrium position.
In this case, the force exerted by the spring can be found as follows:
F = -kx
F = -100 N/m * 0.04 m
F = -4 N
The negative sign indicates that the force is in the opposite direction to the displacement of the mass from its equilibrium position.
Next, we can use the work-energy principle to find the speed of the mass as it passes through its equilibrium position. The work-energy principle states that the work done on an object is equal to its change in kinetic energy. In this case, the work done on the mass is the work done by the spring, which is:
W = (1/2)kx^2
where W is the work done by the spring and x is the displacement of the mass from its equilibrium position.
The work done by the spring can be found as follows:
W = (1/2)kx^2
W = (1/2)(100 N/m)(0.04 m)^2
W = 0.008 J
This work done by the spring is equal to the change in kinetic energy of the mass, which is:
ΔK = (1/2)mv^2 - 0
where ΔK is the change in kinetic energy of the mass, m is the mass of the mass, and v is its speed.
We can solve for v as follows:
ΔK = (1/2)mv^2 - 0
ΔK = (1/2)mv^2
v^2 = 2ΔK/m
v = sqrt(2ΔK/m)
Substituting the values we have found, we get:
v = sqrt(2(0.008 J)/(0.05 kg))
v = sqrt(0.32 m^2/s^2 / 0.05 kg)
v = sqrt(6.4 m^2/s^2)
v = 2.53 m/s
Therefore, the speed of the mass when it passes through its equilibrium position is 2.53 m/s.