An unstretched spring with spring constant 37 N/cm is suspended from the ceiling. A 3.3 kg mass is attached to the spring and let fall. To the nearest tenth of a centimeter, how far does it stretch the spring?
the work stretching the spring equals the change in the gravitational potential energy
1/2 * k * x^2 = m * g * x
37 x^2 = 3.3 * 980 * x = 3234 x
x^2 - (3234/37)x = 0
x = 3234/37
To find how far the spring stretches when the 3.3 kg mass is attached to it, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
Hooke's Law can be written as:
F = -kx
where F is the force applied on the spring, k is the spring constant (37 N/cm), and x is the displacement or stretch of the spring.
First, let's convert the spring constant from N/cm to N/m to be consistent with the other units:
1 cm = 0.01 m
So the spring constant becomes:
k = 37 N/cm * (1 cm / 0.01 m) = 3700 N/m
Now, we can calculate the displacement of the spring.
The force exerted by the spring is the weight of the mass, given by:
F = mg
where m is the mass (3.3 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F = (3.3 kg) * (9.8 m/s^2) = 32.34 N (to two decimal places)
Using Hooke's Law, we can rearrange the equation to solve for x:
x = -F / k
Substituting the values:
x = -(-32.34 N) / (3700 N/m) = 0.0087 m
To the nearest tenth of a centimeter, the spring stretches approximately 0.9 cm when the 3.3 kg mass is attached to it.