Question

A 7.0-kg box slides along a horizontal frictionless floor at 1.7 m/s and collides with a relatively massless spring that compresses 23 cm before the box comes to a stop. (a) How much kinetic energy does the box have before it collides with the spring? (b) Calculate the work done by the spring. (c) Determine the spring constant of the spring.

(a) 80.94 J
(b) -40.47 J
(c) 8,000 N/m
(d) -4,000 N/m

Answer 1

The kinetic energy of the box before collision is 10.145 J. The work done by the spring is -10.145 J, indicating the energy transferred to the spring. The spring constant is calculated to be 383.15 N/m.

Step-by-step explanation:

To solve the student's question about the 7.0-kg box sliding along a horizontal frictionless surface and compressing a spring, we will use concepts from physics pertaining to kinetic energy, work, and spring constants.

Part (a)

The kinetic energy (KE) before collision is calculated using KE = ½mv², where m is the mass and v is the velocity of the box. Thus, KE = ½ × 7.0 kg × (1.7 m/s)² = 10.145 J. This is the kinetic energy the box has before the collision with the spring.

Part (b)

The work done by the spring (W) can be calculated as the negative of the change in kinetic energy. Since the box comes to a stop, all the kinetic energy is transferred to the spring, W = -KE = -10.145 J. This is the work done by the spring.

Part (c)

To find the spring constant (k), we use the formula for the potential energy stored in a compressed spring, PE = ½kx², where x is the compression distance. Equating the potential energy to the work done W and solving for k gives us k = -2W/x² = -2(×-10.145 J)/(0.23 m)² = 383.15 N/m.