If a spring stretches 0.9 m when a 3-kg weight is attached to it, how much will it stretch when a 4-kg weight is attached to it?
Assuming the spring obey's Hooke's law, then
force = k* (stretch distance)
from
3kg wt.=k*0.9m
we determine k=3kg.wt/0.9m=(10/3) kg.wt/m
So for 4 kg wt, we have
4 kg.wt=(10/3)*distance
Solve for distance.
To find out how much the spring will stretch when a 4-kg weight is attached, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the amount of stretch or compression. Mathematically, it can be expressed as:
F = -kx
Where:
F is the force applied to the spring in Newtons (N),
k is the spring constant in Newtons per meter (N/m),
x is the displacement or stretch/compression of the spring in meters (m).
In this case, we are given that the spring stretches 0.9 m when a 3-kg weight is attached to it. We can use this information to calculate the spring constant (k).
First, let's find the force exerted by the 3-kg weight on the spring. We know that the weight (W) is given by:
W = mg
Where:
m is the mass of the weight in kilograms (kg),
g is the acceleration due to gravity (approximately 9.8 m/s^2).
So, for a 3-kg weight:
W = 3 kg * 9.8 m/s^2
W = 29.4 N
Since the force exerted by the spring is equal to the weight attached to it (in the opposite direction), we can say that:
F = 29.4 N
Now, we can use the formula F = -kx. We will use the values for F and x from the previous step to find k:
-29.4 N = -k * 0.9 m
Solving for k:
k = 29.4 N / 0.9 m
k ≈ 32.67 N/m
Now that we have found the spring constant (k), we can use Hooke's Law to find the stretch for a 4-kg weight.
F = -kx
We know that F will be the force exerted by the 4-kg weight, which is:
W = mg
W = 4 kg * 9.8 m/s^2
W = 39.2 N
Using Hooke's Law with the new weight:
39.2 N = -32.67 N/m * x
Solving for x (the stretch):
x = 39.2 N / -32.67 N/m
x ≈ -1.2 m
Considering the negative sign, the spring will stretch approximately 1.2 meters when a 4-kg weight is attached to it.