Two blocks of masses m1 = 4.4 kg and m2 = 4.8 kg are connected by a string as shown in the figure above. Block 1 moves at a constant velocity down the incline 30 degrees) block two is not at incline it is straight. they are attached with a string.
a) Find the coefficient of kinetic friction assuming it is the same for both blocks:
-> I have made acceleration 0 (zero) because velocity is constant, however using the mass of the first block on the incline I tried finding the fk (kinetic fric) then using that I found mk (miu kinetic) ... It didn't work
Please help
Why did the string go to therapy?
Because it had tension issues!
But don't worry, I'll try to help you with your physics problem. To find the coefficient of kinetic friction, we need to consider the forces acting on the blocks.
For block 1 on the incline, the forces acting on it include the gravitational force (mg) acting straight down, the normal force (N) perpendicular to the incline, and the kinetic friction force (fk) opposing the motion. Since it is moving at a constant velocity, the net force in the direction of motion is zero. Therefore, we can set up the following equation:
fk = m1 * g * sin(theta)
where m1 is the mass of block 1, g is the acceleration due to gravity, and theta is the angle of the incline.
For block 2, since it is not on an incline, the only force acting on it is the force of kinetic friction (fk) opposing the motion.
If we assume the coefficient of kinetic friction is the same for both blocks, we can set up the following equation:
fk = m2 * g * mu
where m2 is the mass of block 2, g is the acceleration due to gravity, and mu is the coefficient of kinetic friction.
Since both fk values should be equal, we can equate the two equations:
m1 * g * sin(theta) = m2 * g * mu
Now, you can solve this equation to find the coefficient of kinetic friction (mu).
To find the coefficient of kinetic friction, you can follow these steps:
Step 1: Analyze the forces acting on each block separately.
For Block 1 (on the incline):
- The weight (mg) is acting vertically downwards, with a magnitude of m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
- The normal force (N1) is acting perpendicular to the incline, and its magnitude can be determined by decomposing the weight along the inclined plane, N1 = m1 * g * cos(30°).
- The friction force (f1k) is acting parallel to the incline, opposing the motion, and its magnitude can be determined using f1k = m1 * g * sin(30°) * μk, where μk is the coefficient of kinetic friction.
For Block 2 (horizontal):
- The weight (mg) is acting vertically downwards, with a magnitude of m2 * g.
- The normal force (N2) is acting perpendicular to the horizontal surface, and its magnitude is equal to m2 * g.
- The friction force (f2k) is acting horizontally, opposing the motion, and its magnitude can be determined using f2k = m2 * g * μk, where μk is the same coefficient of kinetic friction as for Block 1.
Step 2: Set up equations based on the forces to solve for the coefficient of kinetic friction.
For Block 1:
- The net force along the incline is zero because the block is moving at a constant velocity. Therefore, the gravitational component along the incline (m1 * g * sin(30°)) is balanced by the kinetic friction force (f1k), resulting in the equation m1 * g * sin(30°) = m1 * g * sin(30°) * μk.
- Simplifying the equation, μk = (m1 * g * sin(30°)) / (m1 * g * sin(30°)) = 1.
For Block 2:
- Since the block is moving at a constant velocity, the net force in the horizontal direction is zero. Therefore, the friction force (f2k) and the external force applied to move the system (through the string) are balanced, resulting in the equation f2k = T (where T is the tension in the string).
- Substituting the expression for f2k, we get m2 * g * μk = T.
- Also, T = m1 * g * sin(30°) + f1k = m1 * g * sin(30°) + m1 * g * sin(30°) * μk (from the equilibrium equation of Block 1).
- Substituting the value of μk = 1 (from the previous step), we have m2 * g = m1 * g * sin(30°) + m1 * g * sin(30°).
- Simplifying, we find m2 = 2 * m1 * sin(30°).
Step 3: Solve for the coefficient of kinetic friction.
Substituting the value of m2 = 4.8 kg and m1 = 4.4 kg, the equation becomes:
4.8 kg = 2 * 4.4 kg * sin(30°).
Simplifying further, sin(30°) = 0.5, resulting in 4.8 kg = 8.8 kg.
This is not true, which indicates that there might be an error in the given information or calculations. Please double-check the values provided and ensure that the calculations are done correctly.
To find the coefficient of kinetic friction in this scenario, you can follow these steps:
Step 1: Draw a free-body diagram for both blocks. Label all the forces acting on each block.
For block 1 (on the incline):
- The weight (mg) acting vertically downward.
- The normal force (N) perpendicular to the incline.
- The friction force (fk) acting parallel to the incline.
For block 2 (not on the incline):
- The weight (mg) acting vertically downward.
- The normal force (N) equal to the weight (mg) since it is not on the incline.
Step 2: Write down the equations of motion for each block separately.
For block 1:
m1 * g * sin(θ) - fk = m1 * a (since the block is moving at a constant velocity, its acceleration is zero)
For block 2:
m2 * g = m2 * a (since the block is not on the incline, it moves down with the same acceleration as block 1)
Step 3: Substitute known values and solve the equations.
Using the given values m1 = 4.4 kg, m2 = 4.8 kg, and θ = 30 degrees, we can substitute them into the equations of motion.
For block 1:
4.4 * 9.8 * sin(30) - fk = 4.4 * 0
(4.4 * 9.8 * 0.5) - fk = 0
2.2 * 9.8 - fk = 0
f k = 21.56 N
Step 4: Use the equation for kinetic friction to find the coefficient of kinetic friction.
The equation for kinetic friction is given by fk = mk * N, where N is the normal force.
Since both blocks are assumed to have the same coefficient of kinetic friction, we can equate the normal forces:
m2 * g = N
Substituting the known values:
m2 * 9.8 = 4.8 * 9.8
N = 47.04 N
Now, substitute this value of N and the calculated value of fk into the equation for kinetic friction:
21.56 N = mk * 47.04 N
Solving for mk:
mk = 0.459
Therefore, the coefficient of kinetic friction for both blocks is approximately 0.459.
If I have this picture right block one on the incline is pulling block two on the level above toward the incline.
There is one force pulling block one down the incline and string tension T holding it back
Fpull = 4.4 g sin 30 = 2.2 g
2.2 g - T = mu * 4.4 g cos 30
I assume that string goes over a frictionless pulley so T pulls forward on bloc 2
T = mu * 4.8 g
so
2.2 g - mu * 4.8 g = mu * 4.4 g * .866
2.2 g = mu g (8.61)
mu = 2.2 / 8.61 = .256