Question

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

Answer 1

The correct answer is A

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


-Anonymous

Answer 2

Answer:

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

Explanation:

Force acting normal to the body $= mg sin(theta)$

substituting the given values in above equation, we get -

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

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

Substituting the given values in above equation, we get -

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

so, the most nearest answer is option A