A 3 m, 250 N/m spring is at the bottom of a 10 m, 20 degree frictionless slope with a 30 m drop off the top of the incline. If a 10 kg wight is placed at the end of the spring when it is compressed 2 m, how fast will the block be traveling just before it hits the ground?
The potential energy stored at the bottom has to equal the final KE
You need to relate the starting position at the end of the spring to the final end of the fall. Work the trig out for the height of the incline, and from that the fall.
PEspring=KEfinal+loss of PEgravity
To calculate the velocity of the block just before it hits the ground, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system will remain constant throughout the motion.
First, let's find the potential energy when the spring is compressed by 2 m. The potential energy stored in a spring is given by the formula:
Potential energy (PE) = 1/2 * k * x^2
Where:
k = spring constant = 250 N/m
x = compression of the spring = 2 m
PE = 1/2 * 250 N/m * (2 m)^2
PE = 500 J
Next, let's find the potential energy when the block is at the top of the incline. The potential energy at an elevated position is given by the formula:
PE = m * g * h
Where:
m = mass of the block = 10 kg
g = acceleration due to gravity = 9.8 m/s^2
h = height of the incline = 30 m * sin(20 degrees)
h = 30 m * sin(20 degrees)
h ≈ 10.207 m
PE = 10 kg * 9.8 m/s^2 * 10.207 m
PE ≈ 1000 J
Now, let's calculate the velocity of the block just before it hits the ground using the principle of conservation of mechanical energy. At the bottom of the incline, all the potential energy will be converted into kinetic energy.
Potential energy at the bottom = Kinetic energy just before hitting the ground
PE = KE
PE = 1/2 * m * v^2
1000 J = 1/2 * 10 kg * v^2
Simplifying the equation:
2000 J = 10 kg * v^2
v^2 = 2000 J / 10 kg
v^2 = 200 m^2/s^2
v = √(200) m/s
v ≈ 14.14 m/s
Therefore, the block will be traveling at approximately 14.14 m/s just before it hits the ground.