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A 10-cm-long spring is attached to the ceiling. When a 2.0kg mass is hung from it, the spring stretches to a length of 15cm.

a. What is the spring constant?

b. How long is the spring when a 3.0 kg mass is suspended from it?

Short Answer

Expert verified

a). The spring constant is 392N/m

b). The spring when a 3.0kg mass it long17.5cm

Step by step solution

01

Step 1. Given Information

Length of spring Lo=10cm=0.1m=seq

mass of load m=2Kg

Stretched length of springs=15cm=0.15m

02

Step 2. Part a). The spring constant K

In this case as the load hangs from the spring, it streches because of the weight of load. So the force acting on the spring is weight of load

From Hooke's law:

(Fsp)s=K∆sK=(Fsp)s∆s∆s=s-seq(Fsp)s=mg=(2kg)×(9.8m/s2)N=19.6NK=19.60.15-0.1N/mK=392N/m

03

Step 3. Part b). To find the length of the spring when mass is increased to 3 Kg

As the spring is the same hence its spring constant will remain the same.

Again from Hooke's law

Fsps=K∆s∆s=FspsKFsps=mg=3×9.8=29.4N∵K=392N/ms-seq=29.4N392N/m=0.075ms=0.075m+0.1m=0.175m=17.5cm

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