A 30 kg box is being pulled with a force of 125 N. The coefficient of static friction between the box and the floor is 0.35. What is the minimum downward force on the box that will keep it from slipping?

The minimum downward force on the box that will keep it from slipping is 63.14 N

Explanation:

Given;

mass of the object, m = 30 kg

applied force, f = 125 N

coefficient of static friction, μ = 0.35

Normal reaction (R) is acting upwards, weight of the box (mg) is acting downwards and the minimum downward force (F) on the box that will keep it from slipping is also acting downwards.

The net vertical forces on the box is given by;

R - mg - F = 0

F = R - mg

Now, determine normal reaction, R

f = μR

R = f / μ

R = 125 / 0.35

R = 357.14 N

Finally, determine the minimum downward force on the box that will keep it from slipping;

F = R - mg

F = 357.14 - (30 x 9.8)

F = 357.14 - 294

F = 63.14 N

Therefore, the minimum downward force on the box that will keep it from slipping is 63.14 N

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )