Suppose you have an object of mass of m = 8003 grams that is imitally at rest at

Question

Suppose you have an object of mass of m = 8003 grams that is imitally at rest at a height K above the ground. At some initial time, the object is dropped straight down and his the ground. Assume the value of the gravititional PE is the potenitial energy of the object KE is the kinetic energy of the object. The total initial energy at rest is: E _ initional = PE _ 4 The total final energy when it contacts the ground is: E _intional = KE _ 4 Assuming no losses, the energy conservation law says the initial and final energies must be the same. Therefore: Etmnoi = Efinai The joule (J) is a unit of energy and can be expressed as Kg.m/5 ^ 2 Conversion factors: 12 inches = 1 foot, 2.54 cm = 1 inch Show your work when converting units. Use the correct number of significant figures in your finaal answers If you are told the height h is 28.5 feet, what is the initial energy (in T) of the object

Explanation / Answer

Let the mass of the object m = 8003 g = 8.003 kg

Let the value of g = 9.81 m/s2

Now the height h = 28.5 feet.

1 foot = 12 inches and 1 inch = 2.54 cm

Height h = 28.5 * 12 * 2.54 = 868.68 cm = 8.6868 m

a)Initial energy is nothing but potential energy here

Now the potential energy is given by

                    P.E = m * g * h

                          = 8.003 * 9.81 * 8.6868

                           = 681.99 J

So the initial energy ( P.E ) is 681.99 J

b)Let v be the velocity of the object when it touches the ground

When it touches the ground , all the potential energy must have been converted to kinetic energy

So initial potential energy is equal to kinetic energy at the moment

Then m * g * h = ½ * m * v2

       velocity v2 = 2 * g * h = 2 * 9.81 * 8.6868

                         = 170.43 m2 / s2

Finally           v = 13.054884 m/s

So the velocity is v = 13.0548 m/s

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