Answer:
6N and 10N
Explanation:
The forces acting along the horizontal axis arw the horizontal force applied and the frictional force.
Frictional force is a force that opposes the horizontal force (moving force)
First we need to get the sum of force along the horizontal as shown
\sum Fx = Horizontal force - frictional force
\sum Fx = 8N - 2N
\sum Fx = 6N
Hence one of the magnitude of the net force exerted on the object is 6N
For the forces along the vertical component;
The forces are the weight, force of gravity(acting downward) and the normal force acting upward
Given
Weight = mass × acceleration die to gravity
Weight = 2×10 = 20N
Normal force = 10N
\sum Fy = (weight)-normal force
\sum Fy = (20)-10
\sum Fy = 20-10 = 10N
Hence the other net force acting on the object is 10N
The required options are 6N and 10N
