A 65 kg object is dropped from rest a distance of 4.3m onto a spring with a stiffness coefficient of 6.2x10^4n/m. How far is the spring compressed?
To find how far the spring is compressed, we need to use the concept of potential energy. The potential energy of an object is given by the equation PE = 1/2 k x^2, where PE is the potential energy, k is the spring constant, and x is the compression (displacement) of the spring.
First, let's calculate the potential energy of the object when it is dropped from a height of 4.3 m. The potential energy can be calculated using the equation PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height.
PE = mgh
= (65 kg)(9.8 m/s^2)(4.3 m)
= 2789.15 J
Now, we can equate the potential energy of the object to the potential energy stored in the spring when it is compressed.
PE = 1/2 k x^2
Rearranging the equation, we get:
x^2 = (2PE) / k
x = sqrt((2PE) / k)
Let's substitute the values into the equation:
x = sqrt((2 * 2789.15 J) / (6.2 x 10^4 N/m))
= sqrt(5578.3 / 6.2 x 10^4)
= sqrt(0.089981)
Calculating the square root:
x = 0.30 m
Therefore, the spring is compressed by approximately 0.30 meters.