# Physics

A 1.8 tonne railway truck moving at a speed of 3m/s up a steady incline comes to a stop after 45m. If the sum of the friction opposing the truck amounts to 80N What is the angle of elevation of the slope?

I found Fg using m*g
THen I tried using v^2=u^2 + 2as to find the acceleration so I could find Fnet and then use that to find the !@#\$%^&plied
Adding that to the friction of 80N (because it is in the same direction).
Then I thought to use trigonometry to solve for the angle only it didn'tt give me the correct answer...

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1. Wt. = 1.8 Tonne.
Vo = 3 m/s.
d = 45 m.
Fk = 80 N.

Mass = 1.8 * 1000kg = 1800 kg.

Fg = m*g = 1800 * 9.8 = 17,640 N.

KE + PE = mg*h-Fk*d
KE + PE = mg*h-80*45
0.5m*Vo^2 + 0 = mg*h-3600
900*9 + 3600 = 17640h
17,640h = 11,700
h = 0.6633 m.

sinA = h/d = 0.6633/45 = 0.01474
A = 0.84o = Angle of elevation.

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posted by Henry
2. WHAT DOES A DISPLACEMENT-TIME BALLON TYPE GRAPH REPRESENTS?

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posted by ANU

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