A spring 25cm long is stretched to 30cm by a load of 100N what will be it's length when stretched by a load of 200N provided the elastic limit is not exceeded. Pls I need d solutions and steps to solve this problem now it is my assignment to be submitted tomorrow
stretches 5 cm for 100 N
so
stretches 10 cm for 200 N
so
25 + 10 = 35 cm
To solve this problem, let's start with Hooke's Law, which states that the force exerted by a spring is directly proportional to its extension (change in length) within the elastic limit.
Hooke's Law equation:
F = k * x
Where:
F is the force applied to the spring,
k is the spring constant (a measure of the spring's strength), and
x is the change in length of the spring.
To find the spring constant k, we need to use the given information that a load of 100N stretches the spring from 25cm to 30cm. We can calculate the change in length (x) first:
x = final length - initial length
x = 30cm - 25cm
x = 5cm = 0.05m (converting cm to meters)
Now, we can find the spring constant (k):
100N = k * 0.05m
k = 100N / 0.05m
k = 2000 N/m
Now that we know the spring constant (k), we can use it to find the final length when a load of 200N is applied.
Let's assume the final length is L.
200N = 2000 N/m * (L - 0.25m)
Simplifying the equation:
L - 0.25m = 200N / 2000 N/m
L - 0.25m = 0.1m
L = 0.1m + 0.25m
L = 0.35m
Therefore, the length of the spring when stretched by a load of 200N (within the elastic limit) will be 0.35 meters or 35 centimeters.
Remember to double-check the calculations and units used in your final answer.
Stretch = 30-25 = 5 cm.
Length = 25 + (200N/100N) * 5cm = 25 + 10 = 35 cm.