Question

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.

Answer 1

Answer:

v = 8.49 m/s rounded off to 3 significant figures

Step-by-step explanation

Using Energy Conservation

U = Tk + Vp (Sum of kinetic Tk Energy and gravitational potential Vp Energy)

1. At height = 3.67 m, the hammer was still or the velocity was negligible; hence, vi = 0 m/s.
2. Ui = 0 + mgh
3. At ground all potential energy is converted to kinetic; hence, Uf = $\frac{1}{2}$mv².
4. Since total energy of the system remains constant we equate Ui to Uf.
5. Ui = Uf
6. mgh = $\frac{1}{2}$mv²
7. After cancelling out masses and making v the subject of the formula we get:

v = $\sqrt[2]{2gh}$ = $\sqrt[2]{2*9.81*3.67}$ = 8.49 m/s rounded off to 3 significant figures