Aspring has a force constant of 1 ã— 105 n/m. how far must it be stretched for its potential energy to be 47 j? answer in units of m.

= 0.03066 m

Work done or energy by a spring or an elastic material is given by the formula;

Potential energy = 1/2 kx²

Where, k is the spring constant and x is the distance stretched.

Therefore;

47 J = 1/2 (100000 n/m) x²

 47 J = 50, 000 x²

x² = 47 /50,000

x² = 0.00094

x = 0.03066 m

Therefore, x = 0.03066 m

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.