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
catapult not usually spring, weight hauled back more often
Sorry to say but all this answers are wrong... The answer supposed to be 44.2m/s. Please you guys should solve again
It not correct it suppose to be 44.2 ms-1
Wrong
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
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
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.