A 14.0 kg sled is pullef with a 28.2 N f9rce at a 51.0° angle, acr9ss ground where uk= 0.124. What is the total force acting on the sled?

Explanation:

A mass of 14kg is pulled with a force of 28.2N

Angle of pulled is 51°

Coefficient of kinetic friction = 0.124.

Total force?

We need to know the frictional force.

The weight of the body is

W = mg

W = 14 × 9.81

W = 137.34 N

Then,

From the attachment

Fnet, y = 0

N - W  + FSinθ = 0

N = W - FSinθ

N = 137.34 N - 28.2Sin51

N = 115.424 N

From frictional law,

Fr = μk•N

Fr = 0.124 ×115.423

Fr = 14.31 N.

Then, the net force in the x direction is

Fnet = FCosθ - Fr

Fnet = 28.2Cos51 - 14.31

Fnet = 3.43 N.

So, the net force on the x direction is 3.43

Extras, we can also find the acceleration

Fnet = ma = 3.43

14a = 3.43

a = 3.43 / 14

a = 0.25 m/s²

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6