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. 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 mv^2 (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?
gh = 12 v^2

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.

Answer 1

1) The mass has the same value before and after the impact (We will neglect the microscopic effects in which there may be a minimum molecular loss of the material) so the equation will be

$mgh = \frac{1}{2} mv^2$

If we put the mass in one side of the expression (or just cancel directly) we have that

$\frac{m}{m} gh = \frac{1}{2} v^2$

$gh = \frac{1}{2} v^2$

2) We must clear from the equation previously found the number two that is dividing the expression (It would go to the other side of the equation to multiply) and the exponent that would pass to the other of the expression as root therefore

$gh = \frac{1}{2} v^2$

$2gh = v^2$

$v = \sqrt{2gh}$

3) We must simply replace the given values, the height is 3.67m and the gravity on the ground is $9.8m / s ^ 2$ therefore

$v = \sqrt{2(9.8)(3.67)}$

$v = 8.4812m/s$

Answer 2

The relation  $gh = \dfrac{1}{2} v^{2}$ describes the conservation of energy and the velocity of hammer down the hill is 8.48 m/s.

Given data:

The falling distance is, h = 3.67 m.

The mass of hammer is, m = 0.55 kg.

(a)

The problem is based on the conservation of energy, which says that at any point the kinetic energy and the potential energy is constant, such that the total energy remains constant.

Since,  the hammer is dropped from rest, the KE (kinetic energy) 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 (potential energy) upon impact is zero. So,

KE at top +PE at top  = KE at bottom + PE at bottom

0 + PE at top = KE at bottom + 0

$mgh = \dfrac{1}{2} mv^{2}\\\\gh = \dfrac{1}{2} v^{2} .......................................................(1)$

Here, g is the gravitational acceleration and v is the velocity of hammer.

(b)

Manipulating the equation (1) as,

$gh = \dfrac{1}{2} v^{2}\\2gh=v^{2}\\v=\sqrt{2gh}$

(c)

Solve for v as,

$v=\sqrt{2gh}\\\\v=\sqrt{2 \times 9.8 \times 3.67}\\\\v=8.48 \;\rm m/s$

Thus, we can conclude that the relation  $gh = \dfrac{1}{2} v^{2}$ describes the conservation of energy and the velocity of hammer down the hill is 8.48 m/s.

Learn more about the conservation of energy here:

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