A weightlifter \"snatches\" a 1200 N weight by exerting a 1400 N average force f

Question

A weightlifter "snatches" a 1200 N weight by exerting a 1400 N average force for the first meter off the ground, then relaxing his grip and "getting under" the bar to catch it and give it a final upward push.
(a) How much work is done in the first 1 m of lifting by the man? By gravity?
(b) What velocity will the weight attain after the one meter lift?
(c) If the man essentially exerts no force starting at 1 m height, how much farther will the bar rise and how long will it take to rise to that height? During that brief time he will finalize his position to "get under" the bar and then push it to full arm extension.
(d) How much additional work must he do to raise the weight to 2.4 m, the height of his full arm extension?

Explanation / Answer

a) work = F x s
by man = 1400 x 1 = 1400 J
by gravity = 1200 x 1 = 1200 J

b) Net work done = 200 J = change in kinetic energy
200 = 1/2 x 1200/9.8 x v2

v= 1.8 m/s

c) v2 = 2 x 9.8 x s

1.82 =2 x 9.8 x s

s = 0.165 m

1.8 = 9.8 x t

t = 0.183 sec

d) Net work to raise 2.4 m = 1200 x 2.4

= 2880 J

Additional work = 2880 - 1400

= 1480 J

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