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