a spring hanging from the ceiling. The spring constant present is 200N/m. If a mass of 50 kg is attached, how far does the spring stretch?
Force=kx
50kg*9.8N/kg=200N/m * x
x= 50*9.8/200 meters
To find how far the spring stretches when a mass is attached, we need to use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement or stretching of the spring.
The formula for Hooke's Law is:
F = -kx
Where:
F is the force applied to the spring
k is the spring constant (in this case, 200 N/m)
x is the displacement or stretch of the spring
Since the spring is hanging vertically, the weight of the attached mass will provide the force applied to the spring. The weight can be calculated using the formula:
Weight = mass * acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s².
Weight = 50 kg * 9.8 m/s² = 490 N
Now we can set up the equation to solve for x:
F = -kx
-490 N = -200 N/m * x
To eliminate the negative signs, we can remove them from both sides of the equation:
490 N = 200 N/m * x
Now we can solve for x by dividing both sides of the equation by 200 N/m:
x = 490 N / 200 N/m
x = 2.45 m
Therefore, the spring stretches by approximately 2.45 meters when a mass of 50 kg is attached to it.