So attached is the correct answer but I have a few answers on how you would obta

Question

So attached is the correct answer but I have a few answers on how you would obtain that. First, isn't Utop supposed to be .5kx^2, which is Elastic Potential Energy because there is a spring? Second I understand how you can set Kinitial to Ufinal because of Principle of Conservation of Energy, but how can you set .5kx^2 equal to all of those? Please explain.

4. A box of mass m is pressed against (but is not attached to) an ideal spring of force constant k, compressing the spring a distance x. After it is released, the box slides up a frictionless incline as shown in Fig. 13.5 and eventually stops. If we repeat this experiment with a box of mass 2m: A) Just as it moves free of the spring, the heavier box will have twice as much kinetic energy as the lighter box. B) Just as it moves free of the spring, the lighter box will be moving twice as fast as the heavier box. C) Box boxes will reach the same maximum height on the incline. D Both boxes will have the same speed just as they move free of the spring. The lighter box will go twice as high up the incline as the heavier box.

Explanation / Answer

from the conservation of energy

the elastic potential energy in the spring is transfered as kinentic enrgy to the box

potential energy of the spring is E=1/2*K*x^2

kinentic enrgy of the box K1=1/2*m*v^2

1/2*k*x^2=1/2*m*v1^2

and this kinetic energy is transferd as potential energy(U1) of of the box as it reaches the maximum

height

1/2*m*v1^2=U1

1/2*m*v1^2=m*g*h1

hence

E=K1=U1

1/2*k*x^2=1/2*m*v1^2=m*g*h1......(1)

now

mass is increased to 2m

E=K2=U2

1/2*k*x^2=1/2*(2m)*v2^2=(2m)*g*h2

1/2*k*x^2=2(1/2*m*v2^2)=2(m*g*h2)....(2)

hence from equation (1) and (2)

K1=2*K2

and

h1=2*h2

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