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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

Respuesta :

levehyt
The correct answer is A

= 1.3 m/s^2; FN = 63.1 N


-Anonymous

Answer:

Option A, a = 1.3 m/s2; FN = 63.1 N

Explanation:

Force acting normal to the body [tex]= mg sin(theta)\\[/tex]

substituting the given values in above equation, we get -

[tex]F = (8 kg) * 9.8\frac{m}{s^2} * sin (50)\\= 60.05 N\\[/tex]

Force in upward direction [tex]= [tex]m* a - 2.4 N[/tex]

Substituting the given values in above equation, we get -

[tex]20 N = 8 *9.8* a - 2.4 N - mg cos(50)a = \frac{72.79}{8*9.8}\\a = 1.0[/tex]

so, the most nearest answer is option A

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