A 2.2 kg bowling ball is dropped vertically from a height of 1.5 m onto a long compression spring (k = 150 N/m) placed vertically underneath it. What is the maximum compression of the spring?
loss of potential energy by ball = m g (1.5 + x)
gain of potential energy by spring = (1/2)(150)x^2
so
2.2 * 9.81 *(1.5+x) = 75 x^2
21.6 x + 32.4 = 75 x^2
or
75 x^2 - 21.6 x - 32.4 = 0
x = [21.6 +/-sqrt(467+9720) ]/150
x = [ 21.6 + 101 ] /150
x = .817 meter
Damon's post is correct.
To find the maximum compression of the spring, we can use the principle of conservation of energy.
1. Calculate the potential energy (PE) of the bowling ball at the initial height before it is dropped.
PE = m * g * h
PE = 2.2 kg * 9.8 m/s^2 * 1.5 m
PE = 32.34 J
2. Calculate the maximum potential energy stored in the spring.
PE_max = (1/2) * k * x^2
where k is the spring constant and x is the maximum compression of the spring.
3. Set the potential energy of the bowling ball equal to the potential energy stored in the spring.
32.34 J = (1/2) * 150 N/m * x^2
4. Solve for x.
x^2 = (2 * 32.34 J) / (150 N/m)
x^2 = 0.4328 m^2
5. Take the square root of both sides to get the value of x.
x = √0.4328 m^2
x ≈ 0.6587 m
Therefore, the maximum compression of the spring is approximately 0.6587 meters.
To find the maximum compression of the spring, we need to use the principle of conservation of mechanical energy. The initial potential energy of the bowling ball is given by mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the height from which the ball is dropped.
The potential energy is converted into the elastic potential energy stored in the spring when it is compressed. The formula for the elastic potential energy in a spring is (1/2)kx^2, where k is the spring constant and x is the displacement or compression of the spring.
Since mechanical energy is conserved, we can set the initial potential energy equal to the elastic potential energy to find the maximum compression of the spring.
Initial potential energy (mgh) = Elastic potential energy ((1/2)kx^2)
Plugging in the given values, we have:
m = 2.2 kg (mass of the ball)
g = 9.8 m/s^2 (acceleration due to gravity)
h = 1.5 m (height)
k = 150 N/m (spring constant)
(2.2 kg)(9.8 m/s^2)(1.5 m) = (1/2)(150 N/m)(x^2)
Simplifying the equation:
32.34 kg*m^2/s^2 = 75 N/m * x^2
Rearranging the equation:
x^2 = (32.34 kg*m^2/s^2) / (75 N/m)
x^2 = 0.4312 m^2
Taking the square root of both sides:
x = +/- 0.656 m
Since the question asks for the maximum compression, we take the positive value:
x = 0.656 m
Therefore, the maximum compression of the spring is 0.656 meters.