A crate with a mass of 1000 kg is being pulled along greased tracks by a winch. The winch is exerting a force of 2000 N in the horizontal direction along the tracks. The coefficient of kinetic friction between the crate and the tracks is 0.2.

20. What is the net force acting on the crate in the horizontal direction?

21.
Using Newton's second law, calculate the crate's acceleration.

Coefficient of Kinetic Friction = 0.2

Mass of the crate = 1000 kg

Force applied = 2000 N

Normal force is the force applied on the object by the surface. It is equal and opposite to the force of gravity

So, we can say that Normal force = | Gravitational Force |

Normal Force = | mg |

Normal Force = 1000 * 9.8

Normal Force = 9800 N

We know that:

coefficient of Kinetic friction = Friction force / Normal force

replacing the known values

0.2 = Friction force / 9800

Friction Force=  0.2 * 9800

Friction Force = 1960 N

We know that a force of 2000 N is being applied on the crate in the Horizontal direction

Frictional force is always opposite to the horizontal force. So, we can say that:

Applied force - Friction Force = Net Force

replacing the variables

2000 - 1960 = Net force

Net Force = 40N

Therefore, a net force of 40N is being applied on the Crate

From newton's second law of motion:

F = ma

replacing the variables

40 = 1000 * a

a = 40/1000

a = 0.04 m/s²

Hence, the crate will have an acceleration of 0.04 m/s²

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

he . if i may ask what i your question exactly?

explanation: