Question

A block with mass 0.400 kg is on a horizontal frictionless sur- face and is attached to a horizontal compressed spring that has force constant k = 200 N/m. The other end of the spring is attached to a wall. The block is released, and it moves back and forth on the end of the spring. During this motion the block has speed 3.00 m/s when the spring is stretched 0.160 m. During the motion of the block, what is its maximum speed?

Answer 1

The maximum speed of the 0.400 kg block after being released from the compressed spring on a frictionless surface is 4.00 m/s, which is calculated by equating the elastic potential energy of the spring to the kinetic energy of the block.

Step-by-step explanation:

The student asked about the maximum speed of a 0.400 kg block attached to a horizontal compressed spring after being released on a frictionless surface. Using conservation of energy, we know that the energy in the spring when it is compressed or stretched is converted to kinetic energy as the block moves. The block has a known force constant k equal to 200 N/m, and a speed of 3.00 m/s when the spring is stretched by 0.160 m.

First, we calculate the elastic potential energy (EPE) in the spring when it is stretched:

EPE = 1/2 k x^2

EPE = 1/2 (200 N/m) (0.160 m)^2

EPE = 2.56 J

At the point of maximum speed, all the elastic potential energy has been converted to kinetic energy (KE), so:

KE = 1/2 m v_max^2 = EPE

v_max = sqrt((2 * EPE) / m)

v_max = sqrt((2 * 2.56 J) / 0.400 kg)

v_max = 4.00 m/s

Therefore, the maximum speed of the block is 4.00 m/s.