Spring constant

Classical Mechanics Level 2

The system in the above diagram is static. Find $x$ (in cm), the extension of the spring, when the spring constant $k$ is $100 \text{ N/m}.$

Gravitational acceleration $g= 10 \text{ m/s}^2.$

The answer is 20.

6 solutions

Ronald Das
Apr 29, 2014

Tension in the rope is equal to 30N ( taking g=10m/s^2), since the system is in equilibrium . Apply force balance on 1kg block => 30=10 +100x X=0.2m i.e 20cm. I have also considered the rope to be mass less, otherwise the tension would have been varying

the net force acting on the spring is 20 N. thus x=F/k so , x=20/100=0.2m. in cm it is 20cm

Rafay Ahmed Tariq - 7 years, 1 month ago

Mayank Holmes
May 31, 2014

force due the 3kg mass= 30N ; downward force on the spring( due to the 1 KG mass)= 10N; therefore on the spring (net) = (30-10)N, since force on a spring= k*x...and k=100N/M solve for x.......

Kartikay Shandil
May 12, 2014

By balancing forces for equilibrium condition we get m1g=T T-tension, m1=3. g=10m/s2 m2g-T=-Kx T-tension,m2=1,k=100(Spring const.)

Solving we get x=0.2m=20cm

Jaivir Singh
May 6, 2014

30 = 10 + 100 x

Angelito Matibag
May 6, 2014

Computing for the net force:

Fnet=9.81(3-2)

Fnet=19.62 N ≈20 N

Since it is static, the elongation is:

δ=Fnet/k

δ=20/100

δ=0.2 m

δ=20cm

Rafay Ahmed Tariq
Apr 30, 2014

the net force acting on the spring is 20 N. thus x=F/k so , x=20/100=0.2m. in cm it is 20cm

Spring constant

The answer is 20.

6 solutions

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