A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The

Question

3 years ago

11

coefficient of kinetic friction between the box and the floor is 0.23. What is the box's launch speed?

1 answer:

Damm [24] · Answer 1 · 2021-12-31T17:55:40+03:00

Damm [24]3 years ago

3 0

Answer:

ugmd = 1/2 kx²

d = (1/2 kx²) / (ugm)

= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)

= 7.4 m

ugmd = 1/2 mv²

v = √2ugd

= √(2(0.23)(9.81 m/s²)(7.4 m)

= 5.8 m/s

Explanation: