A FORCE OF 5N EXTEND A SPRING OF NATURAL LENGHT 0.5m BY 0.01m WHAT WILLTHE LENGHT OF THE SPRING WHEN APPLY FORCE IS 20N
it will be exended four times .01m, so the length will be .5m+4*.01m
Give the answer
A load of 2on on a wire of cross-sectional area
To find the final length of the spring when a force of 20N is applied, we can use Hooke's Law, which states that the force applied on a spring is directly proportional to its extension.
Hooke's Law can be mathematically represented as:
F = k * x
Where F is the force applied on the spring, k is the spring constant, and x is the extension of the spring from its natural length.
To determine the spring constant (k), we can rearrange Hooke's Law as:
k = F / x
In this case, we are given that a force of 5N extends the spring by 0.01m. Using these values, we calculate the spring constant as follows:
k = 5N / 0.01m = 500 N/m
Now, we can use the spring constant to determine the final length of the spring when a force of 20N is applied.
F = 20N
k = 500 N/m
Rearranging Hooke's Law, we find:
x = F / k
Plugging in the values, we get:
x = 20N / 500 N/m = 0.04m
Therefore, when a force of 20N is applied, the length of the spring will be 0.04m.