Consider 40 N-force applied at an angle of 37° above the horizontal which pulls a 5-kg box on a horizontal floor. The acceleration of the box is 3 ms-2. How large a fritional force must be retarding the motion of the box
To find the frictional force, we need to first resolve the 40 N-force into its horizontal and vertical components.
The horizontal component of the force can be found using cosine:
F_horizontal = 40 N * cos(37°) = 32 N
The vertical component of the force can be found using sine:
F_vertical = 40 N * sin(37°) = 24 N
Since the box is on a horizontal floor, the vertical force is balanced by the normal force from the floor, so there is no vertical acceleration.
Now, let's calculate the net horizontal force acting on the box:
F_net = F_horizontal - F_friction
Using Newton's second law (F = m * a), we can set up the equation:
F_net = m * a
32 N - F_friction = 5 kg * 3 m/s^2
Solving for F_friction:
F_friction = 32 N - (5 kg * 3 m/s^2)
F_friction = 32 N - 15 N
F_friction = 17 N
Therefore, a frictional force of 17 N must be retarding the motion of the box.