a body of mass2kg is lying on a rough inclined surface of inclination 30 degree .find the magnitude of the force parellel to the incline needed to make the block move .1)up the incline .2)doun the incline. cofficent of static friction is 0.2
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To find the magnitude of the force parallel to the incline needed to make the block move, we can follow these steps:
1) Up the incline:
When the block is at rest and the force applied is not enough to overcome the static friction, the net force acting parallel to the incline is zero. We need to determine the maximum static friction force (Fs) that opposes the applied force.
The formula for static friction is Fs = μs * N, where μs is the coefficient of static friction and N is the normal force acting perpendicular to the incline.
To find the normal force (N), we need to consider the forces acting on the block. The weight of the block (mg) can be resolved into two components: one perpendicular to the incline (N), and the other parallel to the incline (mg*sinθ). Here, θ is the inclination angle.
N = mg*cosθ
N = 2kg * 9.8m/s^2 * cos(30°)
N = 2kg * 9.8m/s^2 * 0.866
N ≈ 16.93N
Now we can calculate the maximum static friction force (Fs) using the coefficient of static friction (μs = 0.2):
Fs = μs * N
Fs = 0.2 * 16.93N
Fs ≈ 3.39N
Therefore, to make the block move up the incline, a force greater than 3.39N is required.
2) Down the incline:
When the block is moving down the incline, the force applied is not required, as gravity assists the block's motion. We need to determine the force of kinetic friction (Fk) that opposes the motion.
The formula for kinetic friction is Fk = μk * N, where μk is the coefficient of kinetic friction and N is the normal force.
To find the normal force (N), we use the same method as before:
N = mg*cosθ
N = 2kg * 9.8m/s^2 * cos(30°)
N ≈ 16.93N
Now we can calculate the force of kinetic friction (Fk) using the coefficient of kinetic friction (μk = μs = 0.2):
Fk = μk * N
Fk = 0.2 * 16.93N
Fk ≈ 3.39N
Therefore, to make the block move down the incline, a force greater than 3.39N is required, considering that the assistance due to gravity is already accounted for.