Answer:
The speed of the mass when it is 10 cm from its equilibrium point is 0.62 m/s.
Step-by-step explanation:
The total energy of the oscillating mass is conserved. This means that the sum of the kinetic energy and the potential energy of the mass will always be the same.
When the mass is 15 cm from its equilibrium position, its potential energy is maximum. This is because the spring is stretched to its maximum length. The kinetic energy of the mass is zero at this point, because the mass is not moving.
When the mass is 10 cm from its equilibrium position, its potential energy is half of its maximum value. This is because the spring is only stretched halfway to its maximum length. The kinetic energy of the mass is half of the total energy, because the mass is moving at its maximum speed.
Let's say that the speed of the mass when it is 10 cm from its equilibrium point is v. The kinetic energy of the mass at this point is:
KE = 1/2 * m * v^2
The potential energy of the mass at this point is:
PE = 1/2 * k * x^2
where:
• m is the mass of the object (350 g)
• k is the spring constant (64 N/m)
• x is the displacement of the object from its equilibrium position (10 cm)
We know that the total energy of the mass is conserved, so we can set the kinetic energy equal to the potential energy:
1/2 * m * v^2 = 1/2 * k * x^2
We can solve for v:
v = sqrt(k * x^2 / m)
Plugging in the values for k, x, and m, we get:
v = sqrt(64 * 0.1^2 / 0.35) = 0.62 m/s
Therefore, the speed of the mass when it is 10 cm from its equilibrium point is 0.62 m/s.