An amount of work equal to 2.18j is required to compress the spring in a spring-gun. What is the launch speed of a 1.57g marble?
**Solution.**
A = 2.18 J , m = 1.57 g = 1.57 ⋅ 1 0 − 3 k g ; A = 2.18J, m = 1.57g = 1.57 \cdot 10^{-3}kg; A = 2.18 J , m = 1.57 g = 1.57 ⋅ 1 0 − 3 k g ; v − ? v - ? v − ?
By the law of conservation of energy: kinetic energy of the marble is equal the potential energy of the compressed spring:
W k = W p . W_k = W_p. W k = W p .
That the amount of work is required to compress the spring in a spring-gun, then:
W p = A . W_p = A. W p = A .
The kinetic energy of the marble:
W k = m v 2 2 . W_k = \frac{mv^2}{2}. W k = 2 m v 2 .
Finally:
m v 2 2 = A . \frac{mv^2}{2} = A. 2 m v 2 = A . v = 2 A m . v = \sqrt{\frac{2A}{m}}. v = m 2 A . v = 2 ⋅ 2.18 1.57 ⋅ 1 0 − 3 = 52.7 ( m s ) . v = \sqrt{\frac{2 \cdot 2.18}{1.57 \cdot 10^{-3}}} = 52.7 \left(\frac{m}{s}\right). v = 1.57 ⋅ 1 0 − 3 2 ⋅ 2.18 = 52.7 ( s m ) .
**Answer:** v = 52.7 m s v = 52.7 \frac{m}{s} v = 52.7 s m .