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qaws [65]
3 years ago
14

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

Physics
2 answers:
Mkey [24]3 years ago
6 0

As per the question the mass of sled[m] is given as 8 kg.

The fictional force[F_{f} ] is given as 2.4 N

A force[F_{p} ] of 20 N  was exerted on the sled at angle of 50^{0}

Resolving the force into horizontal and vertical components we get -

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

Here F_{h} is the horizontal component and F_{v} is the vertical component.

F_{h} =20*cos50   =12.85575219 N

Similarly F_{v} =20*sin50=15.32088886 N

From the free body diagram we get that sum of vertical components is zero as there is no motion in vertical direction.

Hence F_{N} +F_{p} sin\theta=mg where g is the acceleration due to gravity .

⇒F_{N} =mg-F_{p} sin50

 ⇒ F_{N} =8*9.8-15.32088886  [here value of g is 9.8 m/s^2]

                      =63.1 N

The net motion of the body is along the  forward  direction.

            Hence  F_{p} cos\theta -F_{f} =ma where a is the acceleration of the body.

              ⇒a=\frac{F_{p}cos\theta -F_{f}  }{m}

                     =\frac{12.85575219 -2.4}{8}

                      =1.30 m/s^2

Hence the option A is the right answer.

       



aivan3 [116]3 years ago
4 0
A. a<span> = 1.3 m/s^2</span><span>; </span>FN<span> = 63.1 N</span>
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