Xiaoyu Yan

Student No.: 28162148

Physics 101 Learning Object:

Energy conservation in the model of spring A spring with stiffness of 10N per meter (k=10N/m) has the initial length of 0.5 meter when it is placed vertically on the ground. At the point that a block with mass of 0.2 kg is put on a spring, the spring is compressed and has the length of only 0.1 meter.

Question: What is the total energy when the block is put on the spring as well as the maximum velocity of the block? (Use g=10N/kg to solve the problem.)

Solution: When the spring is at equilibrium state, force on the spring equals to the

weight of the block.F=G=m×g=0.2kg×10N=2.0N

And to calculate the change in length of the spring, we can use the stiffness of spring.F=k× y, y=F/k=2.0(N)/ 10(N/m)=0.2m0.2m is also known as the Amplitude of spring.

Since the energy in the spring is conserved we can think about the condition, which the spring is only 0.1 meters above the ground, to calculate the total energy. (K=0)

E=U+K=U=12

kA2=0.5×10(N /m)× (0.2m )2=0.2J

Also, to calculate the maximum velocity we can think of the condition when the spring is at equilibrium, the potential force of block equals to zero (U=0) but the total energy is still 0.2J due to energy conservation.

E=U+K=K=12mv2=0.2J

So,v2= E0.5m

= 0.2 J0.5×0.2kg

=2Jkg, v=√2m / s