A 126 kg block is released at a 6.8 m height
as shown. The track is frictionless. The block
travels down the track, hits a spring of force
constant k = 1999 N/m.
The acceleration of gravity is 9.8 m/s
2
.moreno (lm35839) – Homework 16A – smith – (14122) 3
6.8 m
x
126 kg
1999 N/m
Determine the compression of the spring x
from its equilibrium position before coming to
rest momentarily.
Answer in units of m
To determine the compression of the spring x before the block comes to rest momentarily, we can use the principle of conservation of mechanical energy.
1. First, let's calculate the potential energy of the block at its initial height. The potential energy is given by the formula PE = mgh, where m is the mass of the block (126 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (6.8 m). So the potential energy is PE = 126 kg * 9.8 m/s^2 * 6.8 m.
2. Next, as the block travels down the track, it loses potential energy and gains kinetic energy. The formula for kinetic energy is KE = (1/2)mv^2, where m is the mass of the block (126 kg) and v is its velocity.
3. When the block reaches the spring, its velocity is zero as it momentarily comes to rest. At this point, all the potential energy is converted into the potential energy stored in the spring. The formula for the potential energy of the spring is given by PE_spring = (1/2)kx^2, where k is the force constant of the spring (1999 N/m) and x is the compression of the spring.
4. Since mechanical energy is conserved, we can equate the initial potential energy to the potential energy stored in the spring. This gives us 126 kg * 9.8 m/s^2 * 6.8 m = (1/2) * 1999 N/m * x^2.
5. Rearranging the equation, we get x^2 = (2 * 126 kg * 9.8 m/s^2 * 6.8 m) / 1999 N/m.
6. Finally, we can solve for x by taking the square root of both sides of the equation: x = sqrt((2 * 126 kg * 9.8 m/s^2 * 6.8 m) / 1999 N/m).
Calculating this expression will give us the value of x, which represents the compression of the spring in meters.