A force of 10N causes aspring to extent by 20mm find the force applied that cause an extension of 5mm
The force applied that causes an extension of 5mm is 2N.
To find the force applied that causes an extension of 5mm in the spring, we can use Hooke's Law, which states that the force applied on a spring is directly proportional to the extension or compression of the spring.
The formula for Hooke's Law is:
F = k * x
Where:
F is the force applied
k is the spring constant
x is the extension or compression of the spring
In this case, we know that a force of 10N causes an extension of 20mm. We can use this information to calculate the spring constant (k).
k = F / x
k = 10N / 20mm
k = 0.5 N/mm
Now, we want to find the force applied (F) that causes an extension of 5mm. We can rearrange the formula to solve for F:
F = k * x
F = 0.5 N/mm * 5mm
F = 2.5 N
Therefore, a force of 2.5 N is required to cause an extension of 5mm in the spring.
To find the force applied that causes an extension of 5mm, we can use Hooke's Law which states that the force applied to a spring is directly proportional to the extension of the spring.
Hooke's Law can be written as:
F = k * x
Where:
F is the force applied to the spring,
k is the spring constant, and
x is the extension of the spring.
In this case, we have the force (F1 = 10N) and the extension (x1 = 20mm). We can use this information to determine the spring constant.
k = F1 / x1
k = 10N / 20mm
Now we can use the spring constant to find the force (F2) that causes an extension of 5mm.
F2 = k * x2
where x2 = 5mm.
F2 = (10N / 20mm) * 5mm
F2 = 0.5 * 5mm
F2 = 2.5N
Therefore, the force applied that causes an extension of 5mm is 2.5N.