8 A 3.7-kg block is released from rest and allowed to slide down a frictionless

Question

8

A 3.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spring is attached to a wall (see figure for problem 69 on page 231). The initial height of the block is 0.5 m above the lowest part of the slide. The spring compresses from the initial length of 49.3 cm down to the maximum compression of 10.7 cm.

What is the spring constant of the spring?

**********

CHEGG,

Please start with (show how you derive from) simple/basic formulas, and please clearly label and show ALL your work. Thank you!

Explanation / Answer

F = –kx. ......hook's law

where the minus sign shows that this force is in the opposite direction of the force that’s stretching or compressing the spring. (k is called the spring constant, which measures how stiff and strong the spring is. x is the distance the spring is stretched or compressed away from its equilibrium or rest position.)

The force exerted by a spring is called a restoring force; it always acts to restore the spring toward equilibrium. In Hooke’s law, the negative sign on the spring’s force means that the force exerted by the spring opposes the spring’s displacement.

In our case we consider the spring is in equilibrium condition. It means addition of spring length and maximum compression.

:- 49.3 + 10.7 = 60 cm = 0.60 m

F = mg = 3.7 × 9.8 = 36.30 N

K = F/X = 36.30 / 0.60 = 60.5 N/m ( negative sign disappear because spring is in equilibrium position )

Related Questions

Navigate

• Browse (All)
• Browse 8
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.