Question Details:
A crate of mass1 on a frictionless inclined plane is attached to another crate, mass2 by a massless rope. The rope passes over an ideal pulley so the mass m2 is suspeneded in air. The plane is inclined at an angle 36.9. Use conservation of energy to find how fast crate m2 is moving after m1 has traveled a distance of 1.4m along the incline, startin from rest. The mass of m1 is 12.4kg and m2 is 16.3kg.
So, I think I am solving for the velocity of m2? I know by using a FBD that the tensions and normal force cancel out and all that is left is gravity acting as an outside force. I am stuck after this! Help to find this solution would be much appreciated
If they are attached, they are moving the same velocity.
The energy balance is that M2 looses Potential energy, M2 gains Potential energy, and both gain KEnergy.
M2*gh-M1*g*1.4sinTheta+1/2 =(M1+M2)v^2
solve for v
To solve this problem using conservation of energy, we need to consider the initial and final states of the system. At the beginning, crate m1 is at rest, so its initial kinetic energy is zero.
Let's denote the velocity of crate m2 as v2 after crate m1 has traveled a distance of 1.4m along the incline.
First, we need to calculate the height difference (Δh) between the starting and ending positions of m1. This can be done using the formula:
Δh = 1.4m * sin(36.9°)
Next, we can calculate the potential energy gained by crate m1 as it travels this height difference:
PE1 = m1 * g * Δh
where m1 is the mass of crate m1 and g is the acceleration due to gravity.
The potential energy gained by m1 is then converted into the kinetic energy of m2:
KE2 = m2 * v2^2 / 2
where m2 is the mass of crate m2.
According to the law of conservation of energy, the potential energy gained by m1 is equal to the kinetic energy gained by m2:
PE1 = KE2
Substituting the formulas for PE1 and KE2, we get:
m1 * g * Δh = m2 * v2^2 / 2
Now we can solve for v2:
v2^2 = 2 * (m1 * g * Δh) / m2
v2 = √[2 * (m1 * g * Δh) / m2]
Substituting the given values, with m1 = 12.4 kg, m2 = 16.3 kg, Δh = 1.4m * sin(36.9°), and g = 9.8 m/s^2:
v2 = √[2 * (12.4 kg * 9.8 m/s^2 * (1.4m * sin(36.9°))) / 16.3 kg]
Now you can calculate the value of v2 using a calculator.
To solve for the velocity of m2 using conservation of energy, we can consider the potential and kinetic energy of the system.
1. Start by finding the gravitational potential energy of m1 at its initial position (before it starts moving). The formula for gravitational potential energy is: PE = mgh, where m is the mass of the object (m1), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the object. In this case, the height h is given by h = 1.4m.
2. Next, find the potential energy of m2 at its initial position when m1 has not moved yet. The potential energy of m2 is also given by PE = mgh, where m is the mass of the object (m2), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the object. Since m2 is suspended in air, its height h is equal to 0.
3. Now, find the kinetic energy of m1 after it has traveled a distance of 1.4m. The formula for kinetic energy is KE = 0.5mv^2, where m is the mass of the object (m1) and v is its velocity. Since m1 starts from rest, its initial velocity is 0, so the initial kinetic energy is 0.
4. Find the potential energy of m1 when it has moved a distance of 1.4m. The potential energy is given by PE = mgh, where m is the mass of the object (m1), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the object. In this case, h can be found using the angle of the inclined plane (36.9°) and the distance traveled (1.4m) as h = 1.4m * sin(36.9°).
5. Now that we have the initial and final potential energy and the initial and final kinetic energy, we can apply the conservation of energy principle. The principle states that the total energy (potential energy + kinetic energy) remains constant in a system with no external forces acting on it.
PE1 + KE1 = PE2 + KE2,
where PE1 and KE1 are the initial potential and kinetic energy, and PE2 and KE2 are the final potential and kinetic energy.
6. With the equation from step 5, substitute the values from steps 1-4 and solve for the final velocity of m2 (v).
m1 * g * 1.4 + 0 = m1 * g * h + 0.5 * m1 * v^2
Plug in the given values: m1 = 12.4kg, m2 = 16.3kg, g = 9.8 m/s^2, and solve for v.
After finding the value of v, you will have the velocity of m2 after m1 has traveled a distance of 1.4m along the incline.