Saturday
March 25, 2017

Post a New Question

Posted by on .

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.

  • physics - ,

    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*Ao*deltaL^2 / 2Lo

    where E is youngs modulus
    delta L is elongation
    Lo is initial length
    Ao is initial cross section area

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question