Physics 2017 JAMB Past Questions - Question 3: A string of length 5m is extended by 0.04m when a load of 0.8kg is suspended at the end .How far will it extend if a force of 16N is applied? (g = 10ms)

Correct Answer

D

Explanation

F=mg = ke 8 =k x 0.04 ------------------ (1) 16 = k x e ---------------------(2)Equ (2) : 16 = eEqu (1) : 8 0.04e =16 x 0.04 = 0.08m 8 16 x 0.04= 0.0To solve this problem, we can use Hooke's Law, which states that the extension of a spring or string is directly proportional to the force applied to it. The formula for Hooke's Law is:

F = k * x

Where:
F is the force applied (in newtons),
k is the spring constant (in newtons per meter), and
x is the extension (in meters).

In this case, we are given that the string extends by 0.04m when a load of 0.8kg is suspended at the end. We can use this information to find the spring constant (k).

First, let's convert the mass to force using the formula F = m * g, where g is the acceleration due to gravity (10 m/s^2):

F = 0.8kg * 10m/s^2
F = 8N

Now, we can use Hooke's Law to find the spring constant (k):

8N = k * 0.04m

Solving for k:

k = 8N / 0.04m
k = 200 N/m

Now that we have the spring constant, we can find the extension when a force of 16N is applied:

16N = 200 N/m * x

Solving for x:

x = 16N / 200 N/m
x = 0.08m

Therefore, the string will extend by 0.08 meters when a force of 16N is applied.