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A 68.0 kg diver falls from rest into a swimming pool from a height of 4.90 m. It takes 1.46 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.

User Krflol
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1 Answer

3 votes

Answer:

456.4 N

Step-by-step explanation:

From conservation of energy, the potential energy is converted to kinetic energy hence

mgh=½mv²

Making v the subject then


v=\sqrt {2gh}

Where g is acceleration due to gravity and h is height whike v is final velocity. Substituting 4.9m for h and 9.81 m/s² for g then


v=\sqrt {2* 9.81* 4.9}=9.80499872514015m/s\approx 9.8 m/s

Change in momentum equals to the impulse.

Impulse, I= Ft

Change in momentum, ∆p= m(v-u)

Ft=m(v-u) making F the subject of formula then


F=\frac {m(v-u)}{t}

Where F is force in Newton, t is time in seconds, m is mass of diver, v and u are the final and initial velocities respectively.

Substituting 68 kg for m, 9.8 m/s for v, 0 m/s for u since it is initially at rest and 1.46 s for t


F=\frac {68(9.8-0)}{1.46}=456.43835616438N\approx 456.4N

User Les Paul
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