Question #94528
A stone of 20 g is projected vertically upward with a catapult which rubber called has a force constant of 72 Nm-1, if the tension in the cord at the point of release is 36N find the velocity of projector of the stone and the maximum height attained by the stone
1
Expert's answer
2019-09-16T09:52:19-0400
x=Fk=3672=0.5 mx=\frac{F}{k}=\frac{36}{72}=0.5\ m

From the conservation of energy:


0.5mv2=0.5kx20.5mv^2=0.5kx^2

The velocity of projector of the stone:


v=xkm=0.5720.02=30msv=x\sqrt{\frac{k}{m}}=0.5\sqrt{\frac{72}{0.02}}=30\frac{m}{s}

The maximum height attained by the stone:


h=v22g=3022(9.8)=46 mh=\frac{v^2}{2g}=\frac{30^2}{2(9.8)}=46\ m


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Comments

Bosede Kolawole
18.01.24, 19:02

I love it, thanks for the answer

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