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?

Answer 1

correct answer is A, 1.3 and 63.1

Step-by-step explanation:

just did it on edgen

Answer 2

1) Acceleration of the sled

The acceleration of the sled is given by the net force acting in the direction parallel to the incline. There are two forces acting along this direction: the component of the weight parallel to the ramp (downward) and the friction (upward). Therefore, the net force acting in this direction is

`F=mg sin \theta- F_f =(8 kg)(9.8 m/s^2)(sin 50^(\circ))-2.4 N=57.7 N`

And the acceleration is given by Newton's second law:

`a=(F)/(m)=(57.7 N)/(8 kg)=7.21 m/s^2`

2) Normal force

The normal force acting on the sled is equal to the component of the weight perpendicular to the incline, therefore:

`N=mg cos \theta=(8 kg)(9.8 m/s^2 )(cos 50^(\circ))=50.4 N`