A light string, connects blocks 1 and 2 of masses m1 and m2. These blocks are further connected to a block of mass M by another light string that passes over a pulley of negligible mass and friction. Blocks 1 and 2 move with a constant velocity v down the inclined plane, which makes an angle with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f.
Express your answers to each of the following in terms of m1, m2, g, , and f.
(b) Determine the coefficient of kinetic friction between the inclined plane and block 1.
(c) Determine the value of the suspended mass M that allows 1 and 2 to move with constant velocity down the plane.
(d) The string between blocks 1 and 2 is now cut. Determine the acceleration of block 1 while it is on the inclined plane.
Actual link if you would like to see the figure:
apcentral . collegeboard . c o m / apc / members / repository / physics _ b _ 0 0 . pdf
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To solve this problem, we need to analyze the forces acting on each block and use Newton's laws of motion.
(b) To determine the coefficient of kinetic friction between the inclined plane and block 1, we need to consider the forces acting on block 1:
1. The gravitational force acting on block 1 is given by Fg1 = m1 * g * sin(Θ), where g is the acceleration due to gravity.
2. The normal force (N1) acting on block 1 is equal to the component of the weight perpendicular to the inclined plane, which is N1 = m1 * g * cos(Θ).
3. The kinetic frictional force acting on block 1, denoted as fk1, can be expressed as fk1 = μk1 * N1, where μk1 is the coefficient of kinetic friction.
Since block 1 moves with constant velocity v down the inclined plane, the acceleration is zero. Therefore, the net force on block 1 is zero:
Net force on block 1 = Fg1 - fk1 = 0
Substituting the expressions for Fg1 and N1, we have:
m1 * g * sin(Θ) - μk1 * m1 * g * cos(Θ) = 0
We can now solve for the coefficient of kinetic friction μk1:
μk1 = tan(Θ)
(c) To determine the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the plane, we need to consider the forces acting on block 2:
1. The gravitational force acting on block 2 is given by Fg2 = m2 * g.
2. The normal force (N2) acting on block 2 is equal to the weight of block 2, which is N2 = m2 * g.
3. The kinetic frictional force acting on block 2, denoted as fk2, can be expressed as fk2 = μk2 * N2, where μk2 is the coefficient of kinetic friction.
Since block 2 moves with constant velocity v down the inclined plane, the net force on block 2 is zero:
Net force on block 2 = Fg2 - fk2 = 0
Substituting the expressions for Fg2 and N2, we have:
m2 * g - μk2 * m2 * g = 0
μk2 = 1
(d) When the string between blocks 1 and 2 is cut, we need to determine the acceleration of block 1 while it is on the inclined plane. In this situation, the frictional force on block 1 is no longer present.
The net force on block 1 is given by:
Net force on block 1 = Fg1 - F_parallel
The force parallel to the inclined plane, F_parallel, is given by:
F_parallel = m1 * g * sin(Θ)
Therefore, the net force on block 1 is:
Net force on block 1 = m1 * g * sin(Θ) - m1 * g * sin(Θ) = 0
Since the net force on block 1 is zero, the acceleration of block 1 is also zero.
To determine the coefficient of kinetic friction between the inclined plane and block 1, we can start by analyzing the forces acting on block 1. Here's how you can do it:
1. Identify the forces: The forces acting on block 1 are the force of gravity (mg) acting vertically downward, the normal force (N1) acting perpendicular to the inclined plane, and the kinetic frictional force (f) opposing the motion.
2. Split the force of gravity: Resolve the force of gravity into two components: one parallel to the inclined plane (mg sin(θ)) and one perpendicular to the inclined plane (mg cos(θ)).
3. Write the equation of motion: Since block 1 moves with constant velocity down the inclined plane, the net force acting on it must be zero. Write the equation of motion using Newton's second law:
ΣF = f + mg sin(θ) - N1 = 0
4. Solve for the normal force: The normal force (N1) is equal to the perpendicular component of the force of gravity (mg cos(θ)):
N1 = mg cos(θ)
5. Substitute the expression for N1: Substitute the expression for N1 into the equation of motion:
f + mg sin(θ) - mg cos(θ) = 0
6. Solve for the coefficient of kinetic friction: Rearrange the equation to solve for the coefficient of kinetic friction (μk):
f = μk N1
Substituting N1 = mg cos(θ), we get:
f = μk mg cos(θ)
μk = f / (mg cos(θ))
Therefore, the coefficient of kinetic friction between the inclined plane and block 1 is μk = f / (mg cos(θ)).
Now, let's move on to determining the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the inclined plane:
To find M, we need to consider the forces acting on block 2. Here's what you can do:
1. Identify the forces: The forces acting on block 2 are the force of gravity (m2g) acting vertically downward, the normal force (N2) acting perpendicular to the inclined plane, and the kinetic frictional force (2f) opposing the motion.
2. Resolve the force of gravity: Resolve the force of gravity (m2g) into two components: one parallel to the inclined plane (m2g sin(θ)) and one perpendicular to the inclined plane (m2g cos(θ)).
3. Write the equation of motion: Since blocks 1 and 2 move with constant velocity, the net force acting on block 2 must be zero. Write the equation of motion using Newton's second law:
ΣF = 2f + m2g sin(θ) - N2 = 0
4. Solve for the normal force: The normal force (N2) is equal to the perpendicular component of the force of gravity (m2g cos(θ)):
N2 = m2g cos(θ)
5. Substitute the expression for N2: Substitute the expression for N2 into the equation of motion:
2f + m2g sin(θ) - m2g cos(θ) = 0
6. Solve for M: Rearrange the equation to solve for the mass M:
2f = Mg
M = 2f / g
Therefore, the value of the suspended mass M that allows blocks 1 and 2 to move with constant velocity down the inclined plane is M = 2f / g.
Next, let's determine the acceleration of block 1 when the string between blocks 1 and 2 is cut:
When the string is cut, block 1 becomes an isolated system and experiences a net force due to the friction on the inclined plane. Here's how you can find the acceleration of block 1:
1. Identify the forces: The only force acting on block 1 now is the kinetic frictional force (f) opposing the motion.
2. Write the equation of motion: Write the equation of motion using Newton's second law:
ΣF = f - mg sin(θ) = ma
3. Solve for the acceleration: Rearrange the equation to solve for the acceleration (a):
f - mg sin(θ) = ma
a = (f - mg sin(θ)) / m1
Therefore, the acceleration of block 1 while it is on the inclined plane is a = (f - mg sin(θ)) / m1.
I hope this helps! Let me know if you have any more questions.