Question

A mobile phone, mass 0.112 kg and 1.7 m off the ground is in free fall. When it hits the ground, calculate the average force acting on the mobile phone if it is brought to a stop within the thickness of its 2 mm thick protective flexible cover.

Answer 1

Step 1: Find speed of mobile phone just before it hits on the ground:

Using 3rd kinematic equation:

V^2 = U^2 + 2*a*h

U = Initial speed of phone = 0 m/s, since in free fall

a = acceleration due to gravity = -g = -9.81 m/s^2 (Assuming downward direction is negative)

h = vertical dosplacement = -1.7 m

So,

V = final speed before hitting on ground = sqrt (0^2 + 2*(-9.81)*(-1.7))

V = 5.775 m/s

Step 2: Now find acceleration required to stop phone within the thickness of it, So

Using 3rd kinematic equation:

V1^2 = V^2 + 2*a1*h1

a1 = acceleration required = ?

h1 = displacement before stopping = 2 mm = -2*10^-3 m

V1 = final speed = 0 m/s

So,

a1 = (V1^2 - V^2)/(2*h1)

a1 = (0^2 - 5.775^2)/(2*(-2*10^-3))

a1 = 8337.66 m/s^2

Step 3:

Using newton's 2nd law:

Average force required to stop mobile phone will be:

F_net = m*a1

m = mass of mobile phone = 0.112 kg

F_net = 0.112*8337.66

F_net = 933.8 N

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