Question

4. A spring is hanging vertically with its bottom at a height of 1 m. We attach a 1 kg mass to the spring, and see that, after we gently lower it and let go, the bottom of the spring is at height of 0.95 m. Draw an energy interaction diagram starting at the moment the mass is attached and ending when the mass comes to rest. What is the value of the spring constant for this spring? Part 1: How many energy systems are changing? O 1 0 3 O 4 O 2

Answer 1

4.

m = mass hanged = 1 kg

$\Delta$x = stretch in the length of the spring = 1 - 0.95 = 0.05

k = spring constant = ?

the spring force balance the weight of the mass hanged

hence

gain in spring energy = loss of gravitational potential energy

(0.5) k $\Delta$x²= mg $\Delta$x

(0.5) k (0.05)² = 1 x 9.8 x 0.05

k = 392 J/m²

Part 1 :

number of energy systems changing energy are 2 . potential energy of the mass and spring potential energy of the spring

Part 2:

Part 3 :

k = 392 J/m²