Answer:
0.09
Step-by-step explanation:
Answer:
4Joules
Step-by-step explanation:
According to Hooke's law which states that extension of an elastic material is directly proportional to the applied force provide that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
K is the elastic constant
e is the extension
If a spring exerts a force of 6 N when stretched 3 m beyond its natural length, its elastic constant 'k'
can be gotten using k = f/e where
F = 6N, e = 3m
K = 6N/3m
K = 2N/m
Work done on an elastic string is calculated using 1/2ke².
If the spring is stretched 2 m beyond its natural length, the work done on the spring will be;
1/2× 2× (2)²
= 4Joules
Please find attached photograph. The answers are 9a^2 and 15ab.
