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phys問題題題

2011-02-10 4:57 am

a spring hangs from a rigid support with a load of 200g attached to its lower end. it is known that a force of 4.0n will extend the spring by 100cm.
the load is depressed by 20mm and then released
determine the total energy of the system.

點解我take eqm mgh=0->h=0....咁如果計佢lowest position既energy答案5同係eqm 既erengy???
lowest postion=1/2kx^2-mgh
=1/2(4)(0.02)^2-(200/1000)(10)(0.02)
eqm=1/2(200/1000)(aw)^2

有d人話...做shm戈陣可以neglect gpe...但點解我做緊一題:
a spring of k=140nm^-1 is suspended vertically with its top end rigidly clamped.a mass of 0.14kg is attached to the lower end of the unstretched spring and released
the mass osc.but after long while it is at rest.determine the % loss in mechanical erengy due to frictional loss.
呢題又要睇gpe????thx

更新1:

我知點解了THX

回答 (1)

天同

2011-02-12 8:45 pm

✔ 最佳答案

I don't thnik you could neglect the gravitational potential energy in the calculation of total energy, as the graviational potential energy does take part in energy conversion when the mass oscillates up and down.
The simplest way to find the total energy is to consider the mass position when the spring is fully extended, i.e. the mass is at its lowest position. At this instant, the gravitational potential energy is zero (because of lowest position), the kinetic energy of the mass is zero (because it is about to turn back). The total energy thus equals to the energy stored in the spring.
In your question, you could not arbitrarily select the equilibrium position as a reference position in calculating the gravitational potential energy. Arbitrarily select a reference position, though would not alter any energy change calculation, surely affects the absolute energy.

Q: 點解我take eqm mgh=0->h=0....咁如果計佢lowest position既energy答案5同係eqm 既erengy???
lowest postion=1/2kx^2-mgh <--- not correct, mgh should be added, as gravitational potential energy is added to the system in the downward motion of the mass

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