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3) A 80Kg sled is being pulled with a force of 150N. The kinetic friction coefficient between the sled
and the snow is.125. What is the magnitude of Acceleration of the sled?

Respuesta :

Answer:

[tex]a = 0.65\ m/s^2[/tex]

Explanation:

Net Force

There are four forces acting on the sled:

Its weight W directed down

The normal force N directed up

The applied force F directed left

The friction force, opposite to the motion, directed right.

The forces are shown in the free-body diagram of the figure below.

The body is in equilibrium in the vertical direction, thus:

N = W = m.g

N = 80 Kg . 9.8 [tex]m/s^2[/tex]

N = 784 Nw

In the horizontal direction, the sum of the force vectors is the net force:

Fn = F-Fr

It's assumed the left as positive.

The friction force is calculated as:

[tex]Fr =\mu_k . N[/tex]

Fr =0.125*784 Nw

Fr = 98 Nw

According to Newton's second law, the net force is equal to the mass by the acceleration:

F-Fr=m.a

The acceleration can be calculated by solving for a:

[tex]\displaystyle a=\frac{F-Fr}{m}[/tex]

[tex]\displaystyle a=\frac{150-98}{80}[/tex]

[tex]\mathbf{a = 0.65\ m/s^2}[/tex]

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