A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incline and has a tension of 140 N in it. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. How fast is the wagon going after moving 80 m up the hill? (THe Answer my teacher gives us is 7.90 m/s, but I have no idea how you get it)

Question

contestada

Respuesta :

lublana lublana · Answer 1 · 2020-02-21T06:55:01+01:00

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

[tex]\theta=18.5^{\circ}[/tex]

Tension=[tex]140 N[/tex]

Initial velocity of wagon=[tex]u=0[/tex]

Displacement=s=80 m

Net force applied on wagon=[tex]F=T-mgsin\theta=140-40(9.8)sin18.5=15.7 N[/tex]

By using [tex]g=9.8m/s^2[/tex]

[tex]a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2[/tex]

We know that

[tex]v^2-u^2=2as[/tex]

Using the formula

[tex]v^2=2\times 0.39\times 80[/tex]

[tex]v=\sqrt{2\times 0.39\times 80}=7.9m/s[/tex]