Final answer:
The work done by a spring force when stretching from 1mm to 2mm is calculated using Hooke's Law to first determine the spring constant, and then applying the work on a spring formula, resulting in 0.45 Joules. thus, the correct option is B).
Step-by-step explanation:
The question pertains to calculating the work done by the spring force when a spring is further extended from 1mm to 2mm. Since the force required to deflect the spring by 2.5mm is 750 N, we can determine the spring constant (k) using Hooke's Law, which is F = kx, where F is the force and x is the displacement.
To find the spring constant, we rearrange the equation to k = F/x, which gives us k = 750 N / 0.0025 m = 300,000 N/m. The work done on a spring is given by W = 0.5 * k * (x2 - x02), where x0 and x represent the initial and final displacements from the equilibrium position, respectively.
So, the work done to stretch the spring from 1mm to 2mm is W = 0.5 * 300,000 N/m * ((0.002 m)2 - (0.001 m)2) = 0.45 J. Hence, the correct answer is b. 0.45 J.
The complete question is: content loaded
If 750 N of force deflect a spring 2.5mm, the spring work required to pull the spring from 1mm to 2mm is _________ J
a 3
b 0.45
c 0.6
d 0.9 is: