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1 month ago

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A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

2.4 N. The free-body diagram shows the forces acting on the sled.
What is the acceleration of the sled and the normal force acting on it, to the nearest tenth?

1) a = 1.3 m/s2; FN = 63.1 N
2) a = 1.6 m/s2; FN = 65.6 N
3) a = 1.9 m/s2; FN = 93.7 N
4) a = 2.2 m/s2; FN = 78.4 N

2 answers:

serg [3.5K]1 month ago

6 0

According to the problem, the sled has a mass of 8 kg.

The frictional force $[F_{f} ]$ is quantified as 2.4 N.

A force $[F_{p} ]$ of 20 N was applied to the sled at an angle of $50^{0}$ .

When resolving the force into both horizontal and vertical components, we arrive at -

$F_{h} =F_{p} cos\theta$ and $F_{v} =F_{p} sin\theta$ .

Here, $F_{h}$ identifies the horizontal component while $F_{v}$ indicates the vertical component.

$F_{h} =20*cos50$ $=12.85575219 N$ .

Likewise, for $F_{v} =20*sin50=15.32088886 N$ .

The free body diagram shows that the sum of the vertical forces amounts to zero since there’s no vertical movement.

Consequently, we can express it as $F_{N} +F_{p} sin\theta=mg$ , where g symbolizes the acceleration due to gravity.

Thus, $F_{N} =mg-F_{p} sin50$ .

Consequently, $F_{N} =8*9.8-15.32088886$ , and we recognize that the value of g is 9.8 m/s^2.

This leads to a normal force of = 63.1 N.

The net movement of the sled is in the forward direction.

Hence, as acceleration of the sled is symbolized as $F_{p} cos\theta -F_{f} =ma$ .

We can conclude that $a=\frac{F_{p}cos\theta -F_{f} }{m}$ .

Finally, $=\frac{12.85575219 -2.4}{8}$ .

Thus, option A is correct.

Keith_Richards [3.2K]1 month ago

4 0

A. a<span> = 1.3 m/s^2</span><span>; </span>FN<span> = 63.1 N</span>

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