A 1.20 block is attached to a spring with spring constant 14.0 . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 50.0 . What is The block's speed at the point where = 0.350 A?

To find the speed of the block when x = 0.350 A, where A is the amplitude of the motion, we can use the principle of conservation of mechanical energy.

1. A vibrating system, like a mass-spring system, conserves mechanical energy as the potential energy is converted to kinetic energy and vice versa. In this case, the mechanical energy of the system will be conserved.

2. The total mechanical energy of the system can be expressed as the sum of the potential energy (PE) and kinetic energy (KE):

E = PE + KE

3. The potential energy of the system is given by the formula:

PE = (1/2) k x^2

where k is the spring constant and x is the displacement from equilibrium.

4. The kinetic energy of the system is given by:

KE = (1/2) m v^2

where m is the mass of the block and v is its velocity.

5. At the initial position, when the block is at rest, the potential energy is at its maximum, and the kinetic energy is zero. So, the total mechanical energy is only potential energy:

E1 = PE = (1/2) k x^2

6. At the position x = 0.350 A, the potential energy is zero, and the total mechanical energy is transferred to kinetic energy:

E2 = KE = (1/2) m v^2

7. Since the mechanical energy is conserved, we can equate E1 and E2:

PE = KE

(1/2) k x^2 = (1/2) m v^2

8. Rearranging the equation, we can solve for the velocity v:

v^2 = (k x^2) / m

v = sqrt((k x^2) / m)

9. Plug in the given values:

k = 14.0 N/m (spring constant)
x = 0.350 A (displacement)
m = 1.20 kg (mass of the block)

v = sqrt((14.0 N/m) * (0.350 A)^2 / 1.20 kg)

10. Calculate the value of v using a calculator:

v ≈ 1.144 m/s

Therefore, the block's speed at the point where x = 0.350 A is approximately 1.144 m/s.