Question

When a 0.32-kg block is suspended on a vertical spring, it causes it to stretch 3.50 cm. If the block is now pulled 8.40 cm below its equilibrium position and released, what is the speed of the block when it is 1.00 cm above the equilibrium position?

a) 0.30 m/s
b) 0.20 m/s
c) 0.40 m/s
d) 0.10 m/s

Answer 1

To find the speed of the block when it is 1.00 cm above the equilibrium position, we can apply the law of conservation of energy. By equating the potential energy at the equilibrium position to the kinetic energy at the given position, we can solve for the speed of the block.

Step-by-step explanation:

To find the speed of the block when it is 1.00 cm above the equilibrium position, we can use the law of conservation of energy. At its equilibrium position, the block has potential energy stored in the spring. When it is 1.00 cm above the equilibrium position, this potential energy is converted into kinetic energy. Therefore, we can equate the potential energy at the equilibrium position to the kinetic energy at the given position.

Let's calculate:

1. Find the potential energy at the equilibrium position using the formula PE = 0.5 * k * d^2, where k is the spring constant and d is the stretch of the spring.
2. Equate the potential energy at the equilibrium position to the kinetic energy at 1.00 cm above equilibrium using the formula KE = 0.5 * m * v^2, where m is the mass of the block and v is the velocity.
3. Solve for v to find the speed of the block.