Final answer:
The subject of this question is Physics. It involves the concepts of mass, spring, and force constant. In this case, a body of mass 0.2 kg is suspended from a spring of force constant 80 Nm−1. Using Hooke's law, we can find that the body is displaced by -2.45 cm from its equilibrium position when suspended from the spring.
Step-by-step explanation:
The subject of this question is Physics. It involves the concepts of mass, spring, and force constant.
In this question, a body of mass 0.2 kg is suspended from a spring of force constant 80 Nm−1. The force constant of a spring measures how stiff or stretchy the spring is. In this case, the force constant is 80 Nm−1, which means that for every meter the spring stretches or compresses, it exerts a force of 80 N.
So, if we want to find the force exerted by the spring on the body, we can use Hooke's law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position.
F = -kx
where F is the force exerted by the spring, k is the force constant, and x is the displacement of the body from the equilibrium position.
In this case, the body is suspended from the spring, so the force exerted by the spring is equal to the weight of the body. We can find the weight of the body using the formula:
Weight = mass x acceleration due to gravity
Weight = 0.2 kg x 9.8 m/s² = 1.96 N
So, the force exerted by the spring is 1.96 N. We can now solve for the displacement x:
1.96 N = -80 Nm−1 x
x = -0.0245 m or -2.45 cm
Therefore, the body is displaced by -2.45 cm from its equilibrium position when suspended from the spring.