Asked by Anonymous

A stone of mass 5g is projected with a rubber catapult. If catapult stretch though a distance of 7cm by an average force of 70n, calculate the velocity of the stone when released
Answered by oobleck
find k using F = kx
the energy stored in the rubber is
E = 1/2 kx^2
all that PE became KE, so
1/2 kx^2 = 1/2 mv^2
plug in your numbers to find v.
Answered by Anonymous
work in = 70 * 0.07 =0.49 Joules
(1/2) m v^2 = 0.49
v^2 = .98 / 0.005 = 196
v = 14 m/s
Answered by Anonymous
catapult not usually spring, weight hauled back more often
Answered by Martic
K:E = P:E K:E=1/2MVsquare and P:E=MGH where force=MG 1/2MVsquare=FH make Vsquare subject of formular Vsquare=2FH/M solve Vsquare=2*70*7/5, Vsquare=980/5 , Vsquare=196 , root both sides , V= 14m/s
Answered by Jaiden
Sorry to say but all this answers are wrong... The answer supposed to be 44.2m/s. Please you guys should solve again
Answered by Ransom
It not correct it suppose to be 44.2 ms-1
Answered by Ransom
Wrong
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