A 40 ton truck loses it brakes and reaches the bottom of a hill with a speed of

Question

A 40 ton truck loses it brakes and reaches the bottom of a hill with a speed of 77 mph.
Fortunately, there is a runaway truck ramp which is inclined at angle of 17 degrees to the horizontal.

Assuming no losses, what distance does the truck travel along the runaway truck ramp?

If instead of there being no losses, there is gravel on the runaway truck ramp that provides a constant force of 3240 pounds in the opposite direction of motion. What distance does the truck travel along the runaway truck ramp in this case?

Explanation / Answer

First lets do some conversions...

40 Tons = 36287 kg

3240 lbs = 14412 N

77 mph = 34.4 m/s

Part A)

With no losses, the initial KE = the final PE

.5mv2 = mgh (mass cancels)

.5(34.4)2 = (9.8)(h)

h = 60.4 m

The distance along the ramp is from the sin function

sin (17) = 60.4/d

d = 206.5 m

Part B)

By the work-energy theorem...

KE = PE + W

.5mv2 = mgh + Fd

.5(36287)(34.4)2 = (36287)(9.8)d(sin 17) + 14412d

2.15 X 107 = 1.23 X 106(d)

d = 181.4 m

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