3. 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)
A. 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?
B. 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 = ?)
C. Use your equation from part B to find the speed with which the hammer struck the ground.