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

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.

       

Answer 2

A. a = 1.3 m/s^2; FN = 63.1 N