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

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

Step-by-step explanation:

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

substituting the given values in above equation, we get -

`F = (8 kg) * 9.8(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 = (72.79)/(8*9.8)\\a = 1.0`

so, the most nearest answer is option A