A box weighing 136. N rests on a table. A rope tied to the box runs vertically upward over a massless frictionless pulley and a weight is hung from the other end. What is the force that the table exerts on the box if the weight hanging the other side of the pulley weighs: 2.46x10¯² N ?

m1•g=N+T

m2•g=T
m1•g=N+ m2•g
N=(m1-m2)g

To find the force that the table exerts on the box, we need to consider the forces acting on the box.

First, let's define the upward direction as positive.

The weight of the box itself, which acts vertically downward, can be calculated by multiplying its mass (m) by the acceleration due to gravity (g):

Weight of the box = m * g

In this case, the weight of the box is given as 136 N, so we have:

136 N = m * g

Since the mass (m) of the box is given in kilograms, we need to convert the weight from newtons to kilograms by dividing by the acceleration due to gravity (g ≈ 9.8 m/s^2):

136 N = m * 9.8 m/s^2

Solving for m:

m = 136 N / 9.8 m/s^2 ≈ 13.9 kg

Now, let's consider the forces on the box when the weight is hung from the other end of the rope.

The tension in the rope pulling upward can be calculated using Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration:

Net force = m * a

Since the box is at rest, the net force is zero. Therefore, the tension in the rope is equal to the weight of the hanging mass:

Tension = Weight of hanging mass

In this case, the weight of the hanging mass is given as 2.46 × 10^-2 N, so the tension in the rope is:

Tension = 2.46 × 10^-2 N

Since the tension in the rope is also the force that the table exerts on the box, we can conclude that the force exerted by the table on the box is:

Force = Tension = 2.46 × 10^-2 N

Therefore, the force that the table exerts on the box is 2.46 × 10^-2 N.