An elastic extend by 0.1cm,when adepimpe load of 20g hang on it,what a additional would be require to force a further extention of 2.0cm
since F = kx,
(F+20)/2.1 = 20/0.1
To solve this problem, we can use Hooke's Law, which states that the extension of a spring 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 to the spring (in Newtons)
- k is the spring constant (measured in N/m or Newtons per meter)
- x is the extension or compression of the spring (in meters)
To find the additional force required to extend the spring by 2.0 cm (or 0.02 m), we need to calculate the spring constant (k) first.
Given:
Extension = 0.1 cm = 0.001 m
Force = 20 g = 0.02 kg
Using Hooke's Law, we can rearrange the formula to solve for k:
k = F / x
k = 0.02 kg / 0.001 m
k = 20 N/m
Now that we have the spring constant, we can calculate the additional force required to extend the spring by 2.0 cm.
Given:
Extension = 2.0 cm = 0.02 m
Using Hooke's Law:
F = k * x
F = 20 N/m * 0.02 m
F = 0.4 N
Therefore, an additional force of 0.4 Newtons is required to force a further extension of 2.0 cm.