cutie.patootie
A 5.5 kg bowling ball initially at rest is dropped from the top of a 12 m building. It hits the ground 1.75 s later. Find the net external force on the falling ball.
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yrelhan4
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s=1/2at^2
you have s and t find a.
net force=ma. plug and solve.
cutie.patootie
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What is s?
yrelhan4
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s-->distance travelled
a--> acceleration due to gravity here
t-->time taken
cutie.patootie
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Okay, thanks:)
yrelhan4
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cheers!
cutie.patootie
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I 431.2N but that isn't an option..
yrelhan4
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43.12 N is the answer.
cutie.patootie
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So this is what I did..
s=12m
a=?
t=1.75s
12=((1.75)^2a)/2
12=(3.0625a)/2
24=3.0625a
a=7.84
netforce=55*7.84
netforce=431.2N
yrelhan4
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Net acceleration. a = 2*12/1.75²
Net force ma = 5.5* 2*12/1.75² = 43.12 N
cutie.patootie
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I get what your saying now. But the only problem is my choices are 120N, 54N, & 0N. But I can rule out 0N automatically
yrelhan4
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hmm. no idea. sorry.
cutie.patootie
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Its okay, thanks for your help though :) I really do appreciate it
yrelhan4
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you're welcome. good luck. :)