Part I: Applications to Physics and Engineering (20 Points) (1) consider a sprin

Question

Part I: Applications to Physics and Engineering (20 Points) (1) consider a spring that has a natural length of 12.0 centimeters. When a force of 30.0 Newtons is hung from the spring, the spring is extended to 15.0 centimeters. (a) Calculate the spring constant. You may express your answer in Newtons per centimeter or in Newtons per meter, as you prefer. much work done in (b) It is then desired to stretch the spring to 20.0 centimeters. How is in extending the spring from 15.0 centimeters to 20.0 centime Express your answer Joules.

Explanation / Answer

Original length = L1 = 12 cm = 0.12 m

Stretched length = L2 = 15 cm = 0.15 m

Stretch = x = L2 - L1 = 0.15 - 0.12 = 0.03 m

F = applied force = 30 N

Using the formula :

F = kx

30 = k (0.03)

k = 1000 N/m

b)

x1 = initial stretch = 15 cm - 12 cm = 3 cm = 0.03 m

x2 = final stretch = 20 cm - 12 cm = 3 cm = 0.08 m

Work done = change in Spring Potential energy

W = (0.5) k x22 - (0.5) k x12

W = (0.5) (1000) ( (0.08)2 - (0.03)2 )

W = 2.75 J

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