a physics question
2017/06/06 04:26
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發問:
a woman weighing 600N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring compressed 1.0cm under her weight. Find the force constant of the spring.I am puzzled about how to find the force constant correctly.If I consider the force, kx=mg ,then k(-0.01)=-600 ------>k=6*10^4... 顯示更多 a woman weighing 600N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring compressed 1.0cm under her weight. Find the force constant of the spring. I am puzzled about how to find the force constant correctly. If I consider the force, kx=mg ,then k(-0.01)=-600 ------>k=6*10^4 N/m But when I consider the energy change , mgh=0.5kx^2 , 600*0.01=0.5k*(0.01)^2----->we find k=1.2*10^5 ,which is double than the k I found by the force approach. Which one is correct and why the two answers would be different
最佳解答:
The first one is correct. The force constant equals to 6 x 10^4 N/m In your second method, you have neglected the kinetic energy of the woman when the spring is being compressed. The spring, in fact, oscillates when the woman steps on it (though in practice, the oscillation is small and usually unobservable). The kinetic energy of oscillation gradually dies down because of damping (i.e. friction) and transformed into heat. Hence, when the woman comes to rest, only half of the original potential energy is transformed to the spring as elastic potential energy. By conservation of energy, (600 x 0.01)/2 = (1/2)k.(0.01)^2 you would get k = 6 x 10^4 N/m
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a physics question發問:
a woman weighing 600N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring compressed 1.0cm under her weight. Find the force constant of the spring.I am puzzled about how to find the force constant correctly.If I consider the force, kx=mg ,then k(-0.01)=-600 ------>k=6*10^4... 顯示更多 a woman weighing 600N steps on a bathroom scale that contains a stiff spring. In equilibrium, the spring compressed 1.0cm under her weight. Find the force constant of the spring. I am puzzled about how to find the force constant correctly. If I consider the force, kx=mg ,then k(-0.01)=-600 ------>k=6*10^4 N/m But when I consider the energy change , mgh=0.5kx^2 , 600*0.01=0.5k*(0.01)^2----->we find k=1.2*10^5 ,which is double than the k I found by the force approach. Which one is correct and why the two answers would be different
最佳解答:
The first one is correct. The force constant equals to 6 x 10^4 N/m In your second method, you have neglected the kinetic energy of the woman when the spring is being compressed. The spring, in fact, oscillates when the woman steps on it (though in practice, the oscillation is small and usually unobservable). The kinetic energy of oscillation gradually dies down because of damping (i.e. friction) and transformed into heat. Hence, when the woman comes to rest, only half of the original potential energy is transformed to the spring as elastic potential energy. By conservation of energy, (600 x 0.01)/2 = (1/2)k.(0.01)^2 you would get k = 6 x 10^4 N/m
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