PHYSICS
JAMB 2008 - Question 15
Physics 2008 JAMB Past Questions - Question 15: A spring of force constant 500Nm-¹ is compressed such that its length shortens by 5cm .The energy stored in the spring is
Choose the correct answers from the options given.
A:
B:
C:
D:
Correct Answer
A
Explanation
Energy stored in a spring= 1/2k e2 = 1/2 x 500 <0.05>2 = 0.625j
The energy stored in a compressed spring can be calculated using the formula for elastic potential energy:
Elastic Potential Energy (U) = (1/2) * k * x^2
Where:
U = Elastic potential energy
k = Force constant (spring constant) in N/m
x = Compression or extension of the spring from its equilibrium position in meters
In this case, you have the following values:
k = 500 N/m
x = 5 cm = 0.05 m (since 1 cm = 0.01 m)
Now, you can plug these values into the formula:
U = (1/2) * 500 N/m * (0.05 m)^2
U = (1/2) * 500 N/m * 0.0025 m^2
U = 0.5 * 500 N/m * 0.0025 m^2
U = 125 N * 0.0025 m^2
U = 0.3125 joules
So, the energy stored in the compressed spring is 0.3125 joules.
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