Answer to Question #261115 in Classical Mechanics for Tetsuya

Question

1. An airplane has a mass of 35,000 kg and a take-off acceleration of 1.20 m/s2. What is the net force accelerating the plane?

2. Let us say that we have a 20-kg load to be pulled by a rope on a horizontal floor. The rope is making an angle of 30 degrees with the horizontal. The tension of the rope is 40 N. If the body starts from rest,

a. find the distance it will move after 5 seconds

b. the velocity after 5 seconds, and

c. the force exerted by the floor on the body 

Answer 1

1)

$$F_{net}=ma=35000\ kg\times1.20\ \dfrac{m}{s^2}=4.2\times10^4\ N.$$

2) (a) Let's first find the acceleration of the load:

$$F_Tcos\theta=ma,$$ $$a=\dfrac{F_Tcos\theta}{m}=\dfrac{40\ N\times cos30^{\circ}}{20\ kg}=1.73\ \dfrac{m}{s^2}.$$

Finally, we can find the distance the load will move after 5 seconds from the kinematic equation:

$$d=v_0t+\dfrac{1}{2}at^2,$$ $$d=0+\dfrac{1}{2}\times1.73\ \dfrac{m}{s^2}\times(5\ s)^2=21.6\ m.$$

(b) We can find the velocity after 5 seconds from the kinematic equation:

$$v=v_0+at=0+1.73\ \dfrac{m}{s^2}\times5\ s=8.65\ \dfrac{m}{s}.$$

(c) We can find the force exerted by the floor on the body as follows:

$$N=mg-F_Tsin\theta,$$ $$N=20\ kg\times9.8\ \dfrac{m}{s^2}-40\ N\times sin30^{\circ}=176\ N.$$