A horizontal force of 100 N pushes a 12-kg block up a frictionless incline that makes an angle of 25 degrees with the horizontal
To determine the acceleration and the normal force acting on the block, we need to break down the external forces acting on the block.
1. Decompose the weight force: The weight force of the block can be separated into two components - the force acting parallel to the incline (mg*sinθ) and the force acting perpendicular to the incline (mg*cosθ).
Weight force parallel to the incline:
F_parallel = mg*sinθ
Weight force perpendicular to the incline:
F_perpendicular = mg*cosθ
Given:
Mass of the block (m) = 12 kg
Angle of the incline (θ) = 25 degrees
Substituting the values, we have:
F_parallel = 12 kg * 9.8 m/s^2 * sin(25 degrees)
F_parallel ≈ 48.84 N
F_perpendicular = 12 kg * 9.8 m/s^2 * cos(25 degrees)
F_perpendicular ≈ 104.65 N
2. Determine the net force acting on the block:
Net force (F_net) is the vector sum of the applied force and the weight force perpendicular to the incline:
F_net = F_applied - F_perpendicular
Given:
Applied force (F_applied) = 100 N
Substituting the values, we have:
F_net = 100 N - 104.65 N
F_net ≈ -4.65 N
Note: The negative sign indicates that the net force is opposite to the applied force.
3. Calculate the acceleration:
The net force acting on the block is equal to the product of its mass and acceleration:
F_net = m * a
Substituting the values, we have:
-4.65 N = 12 kg * a
Solving for acceleration (a):
a = -0.39 m/s^2
Note: The negative sign indicates that the acceleration is opposite to the applied force direction. In other words, the block is moving downwards the incline.
4. Calculate the normal force:
The normal force (F_normal) is the force exerted by the incline perpendicular to the surface.
F_normal = mg*cosθ
Substituting the values, we have:
F_normal = 12 kg * 9.8 m/s^2 * cos(25 degrees)
F_normal ≈ 104.65 N
Thus, the acceleration of the block is approximately -0.39 m/s^2, and the normal force exerted by the incline is approximately 104.65 N.