Physics 2004 JAMB Past Questions - Question 35: If a force of 5CN stretches a wire from 20m to 20.01m, what is the amount of force required to stretch the same material from 20m to 20.05m?

Correct Answer

C

Explanation

K=F/e =50/0.01 =5000N/mF=Ke =5000*0.05=250NTo find the amount of force required to stretch the wire from 20m to 20.05m, you can use Hooke's Law, which relates the force required to the extension of a material within its elastic limit. Hooke's Law is given by the formula:

F = k * ΔL

Where:
F is the force applied (in newtons, N)
k is the spring constant (a material property, specific to the material and its dimensions)
ΔL is the change in length (in meters, m)

In your initial scenario, you have F = 5 N and ΔL = 20.01 m - 20 m = 0.01 m. You can use this information to calculate the spring constant (k):

5 N = k * 0.01 m

Now, solve for k:

k = 5 N / 0.01 m
k = 500 N/m

Now that you have the spring constant, you can use it to find the force required to stretch the wire from 20m to 20.05m:

ΔL = 20.05 m - 20 m = 0.05 m

F = k * ΔL
F = 500 N/m * 0.05 m
F = 25 N

So, the amount of force required to stretch the wire from 20m to 20.05m is 25 newtons.