To solve this problem, we can assume that the spring obeys Hooke's Law, which states that the force applied to a spring is directly proportional to the displacement of the spring from its equilibrium position.
Given that the spring stretches by 6 cm (or 0.06 m) when supporting a load of 15 N, we can use this information to find the spring constant.
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 displacement of the spring.
Plugging in the values, we have:
15 N = k * 0.06 m.
Solving for k, we get:
k = 15 N / 0.06 m
k = 250 N/m.
Now, we can use this spring constant to determine the displacement of the spring when supporting a load of 5 kg (or 49 N).
Using Hooke's Law again, we have:
F = k * x.
49 N = 250 N/m * x.
Solving for x, we get:
x = 49 N / 250 N/m
x = 0.196 m or 19.6 cm.
Therefore, the spring would stretch by 19.6 cm when supporting a load of 5 kg.
A spring stretches by 6cm when supporting a load of 15w.by how much would it stretch when supporting a load of 5kg
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