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Mashutka [201]
3 years ago
13

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 incl

ine 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)
Physics
1 answer:
sp2606 [1]3 years ago
5 0

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

\theta=18.5^{\circ}

Tension=140 N

Initial velocity of wagon=u=0

Displacement=s=80 m

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

By using g=9.8m/s^2

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

We know that

v^2-u^2=2as

Using the formula

v^2=2\times 0.39\times 80

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

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