Asked by karok
the rubber cord of a sling shot has a cross section 2.0 millimetre squared, and an initial length of 20 centimetres. The cord is stretched to 24 centimetres to fire a small stone of mass 10 grams. Assuming that elastic limit is not exceeded, calculate the initial speed of the stone when it is released. (young's modulus(Y)=600000000 newtons per metre squared.
Answers
Answered by
bobpursley
is expression with respect to L:
U_e = \int {\frac{E A_0 \Delta L} {L_0}}\, d\Delta L = \frac {E A_0} {L_0} \int { \Delta L }\, d\Delta L = \frac {E A_0 {\Delta L}^2} {2 L_0}
Potenial energy stored: E*A<sub>o</sub>*deltaL^2 / 2L<sub>o</sub>
where E is youngs modulus
delta L is elongation
Lo is initial length
Ao is initial cross section area
U_e = \int {\frac{E A_0 \Delta L} {L_0}}\, d\Delta L = \frac {E A_0} {L_0} \int { \Delta L }\, d\Delta L = \frac {E A_0 {\Delta L}^2} {2 L_0}
Potenial energy stored: E*A<sub>o</sub>*deltaL^2 / 2L<sub>o</sub>
where E is youngs modulus
delta L is elongation
Lo is initial length
Ao is initial cross section area