dadddynicole
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2. An 82 kg man on a diving board drops from rest 3.0 m above the surface of the water
and he comes to rest 0.55 s after reaching the water. What is the average net force on
the diver as he is brought to rest? Remember to first find the velocity of the man after he
falls 3.0 m off of the diving board.

Answer :

PSN03

Yo sup??

mass of man=82 kg

mgh=1/2mv^2

v^2=2gh

=2*10*3

=60

v=7.74

using v=u+at

0=7.74+a*0.55

a=-14  (approx)

Force=ma

=82*(-14)

=-1148 N

Hope this helps.

adioabiola

The average net force on the man as he is brought to rest is: F = -1143.9N

  • The man on the diving board has a potential energy, P.E = mgh

  • When the man dives, he has a kinetic energy, K.E = (1/2)mv²

Therefore, P.E = K.E

  • mgh = (1/2)mv²

  • 82 × 9.8 × 3 = (1/2)× 82 × v²

  • v² = 58.8

  • V = 7.67 m/s

When the man comes to rest, the final velocity becomes, 0.

As, such, acceleration which is the rate of change of velocity becomes;

  • a = (0 - 7.67)/0.55

  • a = -13.95 m/s².

The force on the man; F = mass × acceleration.

  • F = 82 × -13.95

  • F = -1143.9N

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