A 45.87-kg crate is suspended by a spring with a spring constant of 6000 N/m on a frictionless surface at an angle of 30 °. Find the stretch of the spring in meters.
weight= m g = 45.87 * 9.81 = 450 Newtons
weight component parallel to slope = 450 sin 30 = 225 N
F = k x
225 = 6000 x
x = 0.0375 meters (or 3.75 centimeters or about an inch and a half)
To find the stretch of the spring, we need to calculate the gravitational force acting on the crate and equate it to the force exerted by the spring.
The gravitational force acting on the crate can be calculated using the formula:
F_gravity = m * g
where
m = mass of the crate = 45.87 kg
g = acceleration due to gravity = 9.8 m/s^2 (approximate value)
F_gravity = 45.87 kg * 9.8 m/s^2
F_gravity ≈ 449.37 N
Since the crate is suspended at an angle of 30°, the vertical component of the force exerted by the spring counteracts the gravitational force. The vertical component of the spring force can be calculated using the formula:
F_spring_vertical = F_spring * sin(θ)
where
F_spring = force exerted by the spring
θ = angle of the spring (30°)
To find the force exerted by the spring, we can use Hooke's law:
F_spring = k * x
where
k = spring constant = 6000 N/m
x = stretch of the spring (what we need to find)
Now we can equate the vertical component of the spring force to the gravitational force:
F_spring * sin(θ) = F_gravity
Substituting the values we have:
6000 N/m * x * sin(30°) = 449.37 N
Simplifying:
x * sin(30°) = 449.37 N / 6000 N/m
x * sin(30°) = 0.074895 m
x ≈ 0.074895 m
Therefore, the stretch of the spring is approximately 0.074895 meters.