a 5 kg crate falls from a height of 2.0 m into an industrial spring scale. When the crate comes to rest the compression of the spring is 24 cm. what is the spring constant?
energy stored in spring=PE lost bycrate
1/2 k .24^2=5*9.8*(2.0+.24)^2
solve for k
To find the spring constant, we can use Hooke's Law. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
The formula for Hooke's Law is: F = kx
Where:
F is the force applied to the spring
k is the spring constant
x is the displacement of the spring
In this case, we know the mass of the crate (5 kg), the height it falls from (2.0 m), and the compression of the spring (24 cm).
First, let's convert the compression of the spring from centimeters to meters:
24 cm = 24/100 = 0.24 m
Next, let's calculate the potential energy (PE) of the crate when it falls from a height of 2.0 m:
PE = mgh
where:
m = mass of the crate (5 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = height of the fall (2.0 m)
PE = 5 kg * 9.8 m/s^2 * 2.0 m = 98 J
The potential energy of the crate is equal to the energy stored in the spring when it is compressed. Therefore, we can equate the potential energy to the spring potential energy:
PE = (1/2)kx^2
Solving for the spring constant, we get:
(1/2)kx^2 = PE
kx^2 = 2PE
k = (2PE)/(x^2)
Plugging in the values:
k = (2 * 98 J) / (0.24 m)^2
k = 3437.5 N/m
Therefore, the spring constant is 3437.5 N/m.