A box of mass m=1.5 kg is attached to a spring with force constant k=12 and suspended on a frictionless incline that makes a 30 degree angle with respect to the horizontal. With the spring in its unstretched length, the box is released from rest at x=0. The box slides down the ramp as the spring stretches and momentarily comes to rest at x=xmax.
What is xmax, the maximum extension of the spring when the box is at its lowest point on the incline?
I know the answer is 1.23 but I can't figure out how to get it! I first tried using the forces, setting the force spring equal to the force due to gravity acting on the box, this gave me half of the answer. Should I be using potential energy? I keep getting the wrong answer no matter which equations I use, please help!
What I did on this was mgsintheta = .5kx^2... kept getting it wrong. then realized you have to factor in the distance traveled by the block, that the spring is stretching out to, so what you get is mgsintheta""X"" = .5kx^2 or mgsintheta=.5kx
To find the maximum extension of the spring, we need to consider the forces acting on the box as it moves down the incline.
Let's start by analyzing the forces acting on the box at the lowest point of the incline:
1. Gravitational force: The box experiences a component of its weight acting parallel to the incline, given by mg*sin(30°), where m is the mass of the box and g is the acceleration due to gravity.
2. Normal force: The incline exerts a normal force on the box perpendicular to the incline. This force counteracts the downward component of the weight and is equal in magnitude but opposite in direction to mg*cos(30°).
3. Force due to friction: Since the incline is frictionless, there is no friction acting on the box.
Now let's consider the spring force. When the box is at its lowest point on the incline, it momentarily comes to rest. This means that at this point, the net force on the box is zero.
The net force can be calculated as the sum of the gravitational force component parallel to the incline and the spring force. Since the net force is zero, we have:
mg*sin(30°) - k*xmax = 0
Solving for xmax, we get:
xmax = (mg*sin(30°)) / k
Substituting the known values:
m = 1.5 kg
g = 9.8 m/s^2
k = 12
xmax = (1.5 kg * 9.8 m/s^2 * sin(30°)) / 12
Calculating this expression gives us xmax ≈ 1.226 m
Therefore, the maximum extension of the spring when the box is at its lowest point on the incline is approximately 1.23 m.