JAMB - Physics (1985 - No. 11)
A force of 100N stretches an elastic string to a total length of 20cm. If an additional force of 100N stretches the string 5cm further, Find the natural length of the string
15cm
12cm
10cm
8cm
5cm
Maelezo
Using, f1 = f2
e1 = 20 - L ; e2 = 5cm
\(\frac{f_1}{e_1}\) = \(\frac{f_2}{e_2}\)
\(\frac{100}{20 - L}\) = \(\frac{100}{5}\)
hence, L = 15cm
e1 = 20 - L ; e2 = 5cm
\(\frac{f_1}{e_1}\) = \(\frac{f_2}{e_2}\)
\(\frac{100}{20 - L}\) = \(\frac{100}{5}\)
hence, L = 15cm