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
[tex]F=mg sin \theta- F_f =(8 kg)(9.8 m/s^2)(sin 50^{\circ})-2.4 N=57.7 N[/tex]
And the acceleration is given by Newton's second law:
[tex]a=\frac{F}{m}=\frac{57.7 N}{8 kg}=7.21 m/s^2[/tex]
2) Normal force
The normal force acting on the sled is equal to the component of the weight perpendicular to the incline, therefore:
[tex]N=mg cos \theta=(8 kg)(9.8 m/s^2 )(cos 50^{\circ})=50.4 N[/tex]
Answer:
correct answer is A, 1.3 and 63.1
Explanation:
just did it on edgen
