a block moves up 45degree incline w/ constant speed under the action of a force of 15n applied parallel to the incline.If the coefficient of kinetic friction is 0.3, determine the weight of the block and minimum force required to allow the block to move down the incline at constant speed
To determine the weight of the block and the minimum force required for it to move down the incline at a constant speed, we will use the following steps:
Step 1: Break down the forces acting on the block on the incline:
- The force of gravity, which acts downwards, is known as the weight (W) of the block.
- The normal force (N), which acts perpendicular to the incline.
- The force of friction (f), which acts parallel to the incline.
Step 2: Resolve the weight force into components:
The weight force can be resolved into two components:
- The component parallel to the incline, which opposes the motion and is equal to Wsin(θ), where θ is the angle of the incline (45 degrees in this case).
- The component perpendicular to the incline, which is equal to Wcos(θ).
Step 3: Calculate the normal force:
The normal force (N) is equal to the component of weight perpendicular to the incline, so N = Wcos(θ).
Step 4: Determine the force of friction:
The force of friction (f) can be calculated using the formula: f = μN, where μ is the coefficient of kinetic friction (given as 0.3) and N is the normal force.
Step 5: Calculate the required force to move down the incline:
To determine the minimum force required to move the block down the incline at a constant speed, we need to overcome the force of friction. The minimum force required is equal to the force of friction (f).
Now, let's calculate the weight of the block and the minimum force required step-by-step:
Step 1: Weight of the block:
We are given the force applied parallel to the incline (15 N). This force must balance the force component parallel to the incline due to gravity:
15 N = Wsin(45 degrees)
Solving for W:
W = 15 N / sin(45 degrees)
W ≈ 21.21 N
Therefore, the weight of the block is approximately 21.21 N.
Step 2: Normal force (N):
N = Wcos(45 degrees)
N = 21.21 N * cos(45 degrees)
N ≈ 15 N
The normal force is approximately 15 N.
Step 3: Force of friction (f):
f = μN
f = 0.3 * 15 N
f = 4.5 N
The force of friction is 4.5 N.
Step 4: Minimum force required:
The minimum force required to allow the block to move down the incline at a constant speed is equal to the force of friction:
Minimum force required = 4.5 N
Therefore, the weight of the block is approximately 21.21 N and the minimum force required to allow the block to move down the incline at a constant speed is 4.5 N.
To determine the weight of the block, we can use the formula:
Weight = mass * acceleration due to gravity
Since the block is on an incline, we need to consider only the component of gravity that acts along the incline. The force of gravity can be broken down into two components: one parallel to the incline and one perpendicular to the incline.
The component of gravity that acts parallel to the incline is given by:
Force parallel = Weight * sin(angle)
In this case, the angle is 45 degrees, so we can substitute it into the equation:
Force parallel = Weight * sin(45)
Since we know that the force parallel is equal to the force of kinetic friction, which is 0.3 times the weight of the block, we can set up an equation:
0.3 * Weight = Weight * sin(45)
Simplifying the equation:
0.3 = sin(45)
Now we can solve for the weight of the block:
Weight = 0.3 / sin(45)
Weight ≈ 0.424 N
Therefore, the weight of the block is approximately 0.424 N.
To determine the minimum force required to allow the block to move down the incline at constant speed, we can use the formula:
Force parallel = Force applied - Force of friction
In this case, the force parallel is equal to the force of kinetic friction, which is given as 0.3 times the weight of the block.
Force parallel = 0.3 * Weight
Since the block is moving down the incline at a constant speed, the net force acting on the block must be zero. Therefore, the force applied must be equal to the force of friction:
Force applied = Force of friction
Force applied = 0.3 * Weight
Substituting the value of weight:
Force applied = 0.3 * 0.424
Force applied ≈ 0.1272 N
Therefore, the minimum force required to allow the block to move down the incline at constant speed is approximately 0.1272 N.
A 3.0-kg block moves up a 40° incline with constant speed under the action of a 26-N force acting up and parallel to the incline. What magnitude force must act up and parallel to the incline for the block to move down the incline at constant velocity?
normal force = m g cos 45
so friction down slope = .3 m g cos 45
at 45 sin 45 = cos 45 = sqrt 2/2
weight down slope = m g sin 45
15 N = m g sin 45 + .3 m g cos 45
15 N = (m g sqrt 2/2)( 1 + .3)
30/1.3 sqrt 2 = m g = weight
if friction he other way then
F up = (m g sqrt 2/2)( 1 - .3)
= (m g sqrt 2/2)( .7)