Solve each problem and show all work We will now use energy considerations to find the speed of a falling object at impact. Artiom is on the roof replacing some shingles when his 0.55 kg hammer slips out of his hands. The hammer falls 3.67 m to the ground. Neglecting air resistance, the total mechanical energy of the system will remain the same. The sum of the kinetic energy and the gravitational potential energy possessed by the hammer 3.67 m above the ground is equal to the sum of the kinetic energy and the gravitational potential energy of the hammer as it falls. Upon impact, all of the energy is in a kinetic form. The following equation can be used to represent the relationship: GPE + KE (top) = GPE + KE (at impact) Because the hammer is dropped from rest, the KE at the top is equal to zero. Because the hammer is at base level, the height of the hammer is equal to zero; therefore, the PE upon impact is zero. We may write our equation like this: GPE (top) = KE (at impact) This gives us the equation: (mgh) (top) = 1/2 mv2 (at impact) Notice that the mass of the hammer "m" is shown on both sides of the equation. According to the math rules we have learned, what does this mean? Manipulate the equation (rearrange the variables) to solve for v. (Remember that manipulating an equation does not involve numbers and substitutions. You just rearrange the equation. v = ?) Use your equation from part B to find the speed with which the hammer struck the ground. 8,000,000 kg of water are at the verge of dropping over Niagara Falls to the rocks 50.0 meters below. What is the change in gravitational potential energy as the water splashes on the rocks below?
Solve each problem and show all work
We will now use energy considerations to find the speed of a falling object at impact. Artiom is on the roof replacing some shingles when his 0.55 kg hammer slips out of his hands. The hammer falls 3.67 m to the ground. Neglecting air resistance, the total mechanical energy of the system will remain the same. The sum of the kinetic energy and the gravitational potential energy possessed by the hammer 3.67 m above the ground is equal to the sum of the kinetic energy and the gravitational potential energy of the hammer as it falls. Upon impact, all of the energy is in a kinetic form. The following equation can be used to represent the relationship:
GPE + KE (top) = GPE + KE (at impact)
Because the hammer is dropped from rest, the KE at the top is equal to zero.
Because the hammer is at base level, the height of the hammer is equal to zero; therefore, the PE upon impact is zero.
We may write our equation like this: GPE (top) = KE (at impact)
This gives us the equation: (mgh) (top) = 1/2 mv2 (at impact)
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Notice that the mass of the hammer "m" is shown on both sides of the equation. According to the math rules we have learned, what does this mean?
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Manipulate the equation (rearrange the variables) to solve for v. (Remember that manipulating an equation does not involve numbers and substitutions. You just rearrange the equation. v = ?)
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Use your equation from part B to find the speed with which the hammer struck the ground.
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8,000,000 kg of water are at the verge of dropping over Niagara Falls to the rocks 50.0 meters below. What is the change in gravitational potential energy as the water splashes on the rocks below?
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