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
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.