A catapult used to hold a stone of mass 500g is extended by 20cm with an applied force F. If the stone leaves with a velocity of 40m/s, the value of F is
4.0 x 102N
2.0 x 103N
4.0 x 103N
4.0 x 104N
Explanation
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Since the stone is moving, it is experiencing Kinetic energy and K.E=1/2MV2.
Also recall that Work done by an elastic material is 1/2FE.
Since work done is always equal to energy therefore
Work done by elastic material=K.E of the moving stone.
Therefore
1/2FE = 1/2KE2.
So the halves in both sides will cancel themselves, then what you will have remaining is FE=MV².
Make F subject of the formula.
Then F=(MV²)/E.
M=500g=0.5kg
E= 20cm=0.2m
V=40m/s.
Therefore substituting the values into the equation.
F=(0.5×40×40)÷0.2
F=4000N= 4×10³N.
So myschool is correct

E.P.E=K.E....½Fe=½Mv²....F=?, e=20cm/100---»0.2, M=500g/1000----»0.5, V²=40 x 40---»1600....½ x F x 0.2= ½ x 0.5 x 1600 -----»0.1F=400, F=400/0.1=4000N Or 4 x 10³N

v^2=u^2+2as
s=20cm=0.2m
40x40=0+2x0.2xa
1600/0.4=a
a=4000
f=ma
m=500g..0.5kg
f=0.5x4000
f=2000N=2x10^3N

M= 500g = 0.5kg, v = 40m/s,
e= 20cm = 0.2m, w=?, F=?
W = 1/2mv^2
w= 1/2*0.5 * 40^2
w = 0.25 * 1600
w = 400J
w =1/2Fe
400 = 1/2 * F * 0.2
1/2F = 400/0.2
1/2F = 2000
F = 2000 * 2
F = 4000N
F = 4 * 10^3N
ans = C

Energy stored in extended catapult = Kinetic energy gained by the stone
⇒1/2Fe = 1/2mv²
F = mv²/e
= (0.5 x 40²)/0.2
= 4000
= 4 x 10³ N

what I understand about intense Examination like Jamb, you need to manipulate shortcuts to find a way very fast, not using normal route, you will only end up wasting time. Like this question, I just converted the cm to m and multiply all through to get the answer, no time. 0.5*500*40=4000 which is equal to 4*10^3

myschool please correct your options it's meant to be B not C.
Even the tutorial video says B.
Thank you

i taught 1/2*F*e = 1/2*m*v^2 =1/2*K*e^2
but you people used workdone = force*distance
when we have extension there .
i think the appropiatre formular to use is that of 1/2Force*extension



