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Mechanics help:

Reply 1
Avatar for ShOcKzZ
ShOcKzZ
OP
7
A diver of mass 60kg dives from a diving board at a height of 4m. she hits the water travelling at 8ms^-1 and descends to a depth of 2m in the diving pool. Model the diver as a particle.

a) find the work done against air resistance before the diver hits the water and the average magnitude of the air resistance force.

b) find the average magnitude of the force exerted by the water on the diver as she is brought to rest in the diving pool.

for a) i did G.P.E = 60x9.8x4 = 2352j
K.E - when hittin water = 1/2 x 60 x 64 = 1920
so 2352-1920 = 432j
fxd = 432
f=432/4 = 108

can't do b) ?? any help? thanks..
Reply 2
Avatar for hihihihi
hihihihi
ShOcKzZ
A diver of mass 60kg dives from a diving board at a height of 4m. she hits the water travelling at 8ms^-1 and descends to a depth of 2m in the diving pool. Model the diver as a particle.

a) find the work done against air resistance before the diver hits the water and the average magnitude of the air resistance force.

b) find the average magnitude of the force exerted by the water on the diver as she is brought to rest in the diving pool.

for a) i did G.P.E = 60x9.8x4 = 2352j
K.E - when hittin water = 1/2 x 60 x 64 = 1920
so 2352-1920 = 432j
fxd = 432
f=432/4 = 108

can't do b) ?? any help? thanks..

F=ma it?
Reply 3
Avatar for rts
rts
7
ShOcKzZ
A diver of mass 60kg dives from a diving board at a height of 4m. she hits the water travelling at 8ms^-1 and descends to a depth of 2m in the diving pool. Model the diver as a particle.

a) find the work done against air resistance before the diver hits the water and the average magnitude of the air resistance force.

b) find the average magnitude of the force exerted by the water on the diver as she is brought to rest in the diving pool.

for a) i did G.P.E = 60x9.8x4 = 2352j
K.E - when hittin water = 1/2 x 60 x 64 = 1920
so 2352-1920 = 432j
fxd = 432
f=432/4 = 108

can't do b) ?? any help? thanks..


b) v^2 = u^2 + 2as

a = -16m/s

magnitude of f=ma = 16 x60 = 960N
Reply 4
Avatar for ShOcKzZ
ShOcKzZ
OP
7
hihihihi
F=ma it?

tried it ..probably did it wrong though any suggestions?
Reply 5
Avatar for ShOcKzZ
ShOcKzZ
OP
7
rts
b) v^2 = u^2 + 2as

a = -16m/s

magnitude of f=ma = 16 x60 = 960N

exactly what i did, but the answer is 1548..
Reply 6
Avatar for hihihihi
hihihihi
ShOcKzZ
exactly what i did, but the answer is 1548..

ah u must 60x9.8+960=1548N
Reply 7
Avatar for ShOcKzZ
ShOcKzZ
OP
7
hihihihi
ah u must 60x9.8+960=1548N

ahh right..thanks.
Reply 8
Avatar for flatcaps
flatcaps
I get 1548 too, but you can do it by energy consideration.

When she hits the water she has 1/2 * 60 * 8^2 = 1920 J of kinetic energy.
In decelerating to rest, she therefore loses 1920 J KE which goes into work done against the water.

Also, you've got a PE drop as she descends 2m. mgh = 60*9.8*2 = 1176 J. This work is also done against the water.

Total work done = 1920 + 1176 = 3096 J
W= Fd ==> F = W/d so F = 3096/2 = 1548 N.

Tim
Reply 9
Avatar for john !!
john !!
14
Remember, if you use g=9.8, never write answers to more than 2s.f. :cool:

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