A light spring with spring constant 1.30 10 3 N/m hangs from an elevated support

Question

A light spring with spring constant 1.30 103 N/m hangs from an elevated support. From its lower end hangs a second light spring, which has spring constant 1.95 103 N/m. A 1.50-kg object hangs at rest from the lower end of the second spring.

(a) Find the total extension distance of the pair of springs.
_______m

(b) Find the effective spring constant of the pair of springs as a system. We describe these springs as being in series. Hint: Consider the forces on each spring separately.
____________ N/m

Explanation / Answer

let the extension in the first spring be a and that of the second spring be b

k1 = 1.3 x 103 N/m k2 = 1.95 x 103 N/m

a.) The force exerted by both the springs should be equal since they are connected and it should be equal to the weight of the suspended mass = mg

So, k1 a = k2b = mg

a = mg /k1 = 1.5 x 9.81 / 1300 = 0.01131923 m

b = mg / k2 = 1.5 x 9.81 / 1950 = 0.007546153846 m

So, net extension = a+ b = 0.018865383 m

b.) Let the effective spring constant be k which gives the same extension a+b for the given suspended mass

k (a+b) = mg

k = mg/(a+b) = 1.5 x 9.81 / 0.018865383 = 780 N/m

verify: the relation for effectice spring constant when springs are connected in series is -

1/k = 1/k1 + 1/k2 + 1/k3 + .....

So, 1/k = 1/1300 + 1/1950

k = 1300x1950/(1300+1950) = 780 N/m   

Hence, verified.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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