Thursday, November 16, 2017

Physics in Dropping a Phone

One morning, I unfortuently dropped my phone from my bed. I have a top bunk in a bunk bed, which is about 6 ft, or 1.83 meters, above the floor. Assuming an initial velocity of zero and neglecting air resistance, I calculate the velocity of which my phone hit the ground to be 5.99 m/s.
Vf2 = Vi2 + 2ad
Vf2 = 0 + 2 (9.8 m/s) (1.83 m)
Vfground= 35.9 m2/s2 
Vf= 5.99 m/s

Furthermore, my phone is an iPhone 5S which, according to Apple, has a mass of 112 grams. Setting my dorm room floor at a height of zero, my phone had 2.00 J of energy being converted from potential to kinetic energy throughout its fall. 

Total Energy = K.E + P.E
Total Energy = .5mv2 + mgh
Total Energy = .5 (.112 kg) (5.9 m/s)2   
Total Energy = 2.00 J

Luckily, my screen did not shatter when my phone dropped. There are many factors that play a part into whether a phone will shatter when dropped. The impulse equation, Ft=p is useful in determining this. The higher the force, the more likely it will shatter. Thus the height from which the phone was dropped and the surface it lands on will effect the likelihood of it shattering. The higher the phone is dropped, the larger the momentum and force are. The surface the phone lands on also plays a role in the likelihood of the phone cracking because it effects the time of the collusion. A larger time will occur when landing on something like a pillow and will exert a smaller force. A smaller time which occurs from landing on a surface like tile will exert a larger force. Another equation that determines whether the phone will shatter is P= F/A. The larger the area that ultimately hits the ground, the less pressure exerted on the phone and the less likely it is to shatter.

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