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| # h1psp1 | |
|
A 150 g. mass is suspended from a vertical spring and descends a distance of 4.6 cm, after which it hangs at rest. An additional 500 g. mass is then suspended from the first. What is the total extension of the spring? (neglect the mass of the spring.) |
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| Strategy - the rule "Must have" of Mr.Zhang ® | |
| Find the value of k ; Fs=kx=mg | → (1) |
| And later find x(total stretch length) from ; kx=(m1+m2)g | → (2) |
| Solution |
Find k(constant of spring) from equation(1)
|
| k | = |
m1g
x1
|
| = |
(0.15 kg)(9.8 m·s-2)
0.046 m
|
|
| = | 32 N·m-1 |
Find x(total extension of the spring) from equation(2)
|
| x | = |
(m1+m2)g
k
|
| = |
(0.15 kg+0.50 kg)(9.8 m·s-2)
32 N·m-1
|
|
| = | 0.20 m(or 20 cm.) |
| Ans. The total extension of spring is 20 cm |
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