please help! physics problem!?
A 1800-kg pile driver is used to drive a steel beam into the ground. The pile driver falls 6.00 m before coming into contact with the top of the beam, and it drives the beam 14.0 cm farther into the ground as it comes to rest. Using energy considerations, calculate the average force the beam exerts on the pile driver while the pile driver is brought to rest.
Comments
IN A CERTAIN STEAM HAMMER WHOSE STROKE IS 600MM; THE WEIGHT OF THE TUP IS 360KG.
WHEN THE TUP IS ALLOWED TO FALL FREELY, IT IS FOUND THAT THE AVERAGE INDENTATION PRODUCED IN A PIECE OF HOT IRON IS 6MM.WHAT IS THE AVERAGE FORCE OF THE BLOW.
600 mm = 0.6 metres
360 kg falls 0.6 metres so the PE "lost" = mgh = 360 x 9.81 x 0.6 J
PE "lost" = 2 118.96 Joules
KE "gained" = 2 118.96 = 0.5mv^2 so v ~ 3.431 m/s
You could use v^2 = u^2 + 2as to give v^2 = 0 + 2(9.81)(0.6)
with v ~ 3.431 m/s
For the "tup" we have
s = 6/1000 metres
u = 3.431
v = 0
a = A
t = T
v^2 = u^2 + 2as gives 0 = 3.431^2 + 2(A)(6/1000)
A = -980.98 m/s^2
So the force is F = ma = 360 x 980.98 ~ 353 152 N
You could use WD by the "tup" on the hot iron = Fs = change in KE
F(6/1000) = 2 118.96 so F ~ 353 160 N
Even neater is:
WD by the "tup" on the hot iron = Fs = change in PE
F(6/1000) = 360 x 9.81 x 0.6 so F ~ 353353 160N
That's it...Hope one of the methods is OK.
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