Question

If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its
kinetic energy right before it hits the ground?
J
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Answer 1

The kinetic energy of the cell phone before it hits the ground given the data is 9 J

Data obtained from the question

• Mass (m) = 0.08 Kg
• Velocity (v) = 15 m/s
• Kinetic energy (KE) =?

How to determine the kinetic energy

The kinetic energy of the cell phone can be obtained as follow

KE = ½mv²

KE = ½ × 0.08 × 15²

KE = 0.04 × 225

KE = 9 J

Thus, the kinetic energy of the cell phone is 9 J

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Answer 2

The kinetic energy right before it hits the ground is 0.6 J

Kinetic Energy

Kinetic energy is the energy possessed by a body in motion. The S.I unit is Joule. Kinetic energy depends on:

• The mass of the body

• The velocity of the body

Given that a 0.08 kg cell phone falls off a table at 15 m/s, its

kinetic energy right before it hits the ground can be calculated by using the formula;

K.E = 1/2 x m x $v^{2}$

The given parameters are:

mass m = 0.08 kg

velocity v = 15 m/s

Substitute the parameters into the formula

K.E = 1/2 x 0.08 x 15

K.E = 0.6 J

Therefore, its kinetic energy right before it hits the ground is 0.6 J

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