How Do I Solve Parts C and D for a Stone on a Spring Problem?

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To solve parts c and d of the problem, use the conservation of energy principle. The change in gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height is equal to the elastic potential energy of the spring, which is 70.8 J. For part d, set the initial kinetic energy at the release point equal to the gravitational potential energy at maximum height to find the height. The initial kinetic energy can be expressed as 1/2mv^2, and since the speed at the release point is zero, this simplifies the calculations. By equating the energies, you can determine the maximum height of the stone from the release point.
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I got parts a and b but am having trouble getting part c and d. can someone please help me? thanks in advance!

Figure 8-40 shows an 8.00 kg stone resting on a spring. The spring is compressed 8.0 cm by the stone.

Fig. 8-40

(a) What is the spring constant?
9.8 N/cm by mg/x(compression)

(b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?
70.8 J by convert A to N/m then convert compression to m and then use .5kx^2 to get the answer

(c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height?
J gravitational potential energy is mgy but i don't know y

(d) What is that maximum height, measured from the release point?
m y=(.5mv^2)/mg I know i have to use this eqn. but i don't know v.
 
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parts c) and d)

(c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height?
*** as you said, J gravitational potential energy is mgy but i don't know y, but from conservation of energy you know that dU = EPinitial (see answer B)...

(d) What is that maximum height, measured from the release point?
m y=(.5mv^2)/mg I know i have to use this eqn. but i don't know v.

** actually you don't have to use this equation... we know dU = mg dy = EPinitial... and now you can solve for height..
 


For part c, you can use the conservation of energy principle. At the release point, all the elastic potential energy of the spring is converted into kinetic energy of the stone. At the maximum height, all the kinetic energy is converted into gravitational potential energy. Therefore, the change in gravitational potential energy is equal to the elastic potential energy of the spring at the release point. You can use the answer from part b, 70.8 J, as the change in gravitational potential energy.

For part d, you can use the equation for conservation of energy again. At the maximum height, all the kinetic energy is converted into gravitational potential energy. Therefore, you can set the initial kinetic energy (from the release point) equal to the final gravitational potential energy (at the maximum height) and solve for the height. The initial kinetic energy can be found using the equation for kinetic energy, 1/2mv^2. You can use the mass of the stone and the speed of the stone at the release point (which is zero) to find the initial kinetic energy. Once you have the initial kinetic energy, you can set it equal to the final gravitational potential energy, mgh, and solve for h. This will give you the maximum height of the stone from the release point.
 
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