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?
To determine the speed at which the block will be traveling just before it hits the ground, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system (block + spring) will remain constant throughout the entire process.
Let's break down the problem step by step:
1. Find the potential energy of the block when it is compressed 2 m:
The potential energy in the spring is given by the formula: PE = (1/2)kx²
where PE represents potential energy, k is the spring constant, and x is the compression/extension of the spring.
In this case, the potential energy is:
PE = (1/2)(250 N/m)(2 m)² = 500 J
2. Calculate the gravitational potential energy of the block at the starting point (top of the incline):
The gravitational potential energy is given by the formula: PE = mgh
where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the slope.
In this case, the height of the slope is 10 m. However, since the block is not initially at ground level, we need to consider the height from the ground:
h = 10 m - 30 m = -20 m (negative sign indicates below the starting point)
Therefore, the gravitational potential energy is:
PE = (10 kg)(9.8 m/s²)(-20 m) = -1960 J
3. Use the conservation of mechanical energy to find the final kinetic energy of the block just before it hits the ground:
Since the total mechanical energy is conserved, the sum of the initial potential energy and initial kinetic energy will be equal to the final kinetic energy of the block.
The initial kinetic energy is zero since the block is initially at rest.
Therefore, the final kinetic energy is equal to the sum of potential energies:
KE = 500 J + (-1960 J) = -1460 J
4. Calculate the final velocity using the kinetic energy formula: KE = (1/2)mv²
We have the value of KE (-1460 J) and the mass of the block (10 kg).
Rearranging the formula to solve for velocity, we get:
v = sqrt((2 * KE) / m)
Plugging in the values, we have:
v = sqrt((2 * -1460 J) / 10 kg)
v ≈ 17.1 m/s
Therefore, the block will be traveling at approximately 17.1 m/s just before it hits the ground.