A 6.95 kg bowling ball is dropped 2.20 m onto a large spring. If the bowling ball compresses the spring by 0.55 m, what must the spring constant of the spring be?
total drop = 2.2+ .55 = 2.75 m
drop in gravity Potential energy =
= 6.95 * 9.81 * 2.75
= increase in spring Pe
= (1/2)k x^2 = .5 *k * .55^2
so
k = (6.95*9.81*2.75)/(.5*.55*.55)
To find the spring constant of the spring, you can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position.
Hooke's Law formula is given by:
F = k * x
Where:
F is the force exerted by the spring,
k is the spring constant,
x is the displacement of the spring from its equilibrium position.
In this case, we are given the mass of the bowling ball (m = 6.95 kg), the distance it falls (d = 2.20 m), and the compression of the spring (x = 0.55 m).
First, let's calculate the potential energy (PE) of the bowling ball when it is dropped:
PE = m * g * h
Where:
m is the mass of the bowling ball,
g is the acceleration due to gravity (9.8 m/s^2),
h is the height from which the ball is dropped.
PE = 6.95 kg * 9.8 m/s^2 * 2.20 m
PE = 151.924 J (rounded to three decimal places)
Since potential energy is converted into spring potential energy, we can equate the two:
PE = (1/2) * k * x^2
Rearranging the formula, we get:
k = (2 * PE) / x^2
Plugging in the values, we find:
k = (2 * 151.924 J) / (0.55 m)^2
k ≈ 805.141 N/m (rounded to three decimal places)
Therefore, the spring constant of the spring must be approximately 805.141 N/m.