2. If, in the previous question, there is a 15 N kinetic frictional force opposing the motion, how much work is

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NARA [144]

1) The net force is 16 N to the right

2) The net force is 98 N to the left

3) The net force is 0.5 N downward

4) The net force is 170 N to the right

5) The net force is 175 N to the right

Explanation:

1)

To find the net force, we have to analyze all the forces acting on the box.

We have:

Force to the right: $F_a = 20 N$ , the applied force
Force to the left: $F_f = 4 N$ , the force of friction
Force to the bottom: $F_g = 400 N$ , the weight of the box (the weight is always downward vertically)
Force to the top: $F_N = 400 N$ . This is the normal force, which is the reaction force exerted by the table on the box: it points upward and counterbalances the weight of the box, preventing it from falling down)

Therefore, the horizontal net force is

$F_x = F_a - F_f = 20 - 4 = 16 N$ (to the right)

While the vertical force is

$F_y = F_N - F_g = 400 - 400 = 0$

So the net force is 16 N to the right.

2)

In this case, we have the following forces:

$F_g = 4 N$ downward, the weight of the ball
$F_a = 100 N$ to the left, the force that kicks the ball
$F_f = 2 N$ to the right, the force of friction
$F_N = 4 N$ upward, the normal reaction exerted by the field on the ball

Therefore, the horizontal net force is

$F_x =F_a - F_f = 100 -2 = 98 N$ (to the left)

While the vertical force is

$F_y = F_g - F_N = 4 - 4 = 0$ (downward)

And so, the net force is 98 N to the left.

3)

The force acting on the squirrel in this problem are:

$F_g = 8 N$ downward, the weight of the squirrel
$F_f = 7.5 N$ upward, the air resistance, acting upward

Both forces act vertically and there are no other forces acting in other directions, therefore the net force on the squirrel is simply equal to the net force on the vertical direction, which is:

$F_y = F_g - F_f = 8 - 7.5 = 0.5 N$

And since the weight is larger than the air resistance, the direction of the net force is downward.

4)

The forces acting on Monkey are:

$F_1=95 N$ is the force applied to the right by Bunny
$F_2 = 75 N$ is the force applied by Deer from the left (so, also on the right)
$F_g = 50 N$ is the weight of Monkey, downward
$F_N = 50 N$ is the normal reaction exerted by the surface, upward

So, the net force in the horizontal direction is

$F_x = F_1 + F_2 = 95+75=170 N$ (to the right)

While the net force in the vertical direction is

$F_y = F_N - F_g = 50 - 50 = 0$

And therefore the net force is 170 N to the right

5)

The forces acting on Deer are:

$F_a = 100 N + 100 N = 200 N$ to the right, the combined force applied by Bunny and Monkey
$F_f = 25 N$ to the left, the force of friction
$F_g = 150 N$ downward, the weight of the deer
$F_N = 150 N$ upward, the normal reaction from the surface that balances the weight