- #1
Eclair_de_XII
- 1,083
- 91
Homework Statement
"In Fig. 8-33, a runaway truck with failed brakes is moving downgrade at ##130\frac{km}{h}## just before the driver steers the truck up a friction-less emergency escape ramp with an inclination of ##θ=15°##. The truck's mass is ##1.2⋅10^4kg##. What minimum length L must the ramp have if the truck is to stop momentarily along it?
Homework Equations
##v=130\frac{km}{h}=\frac{325}{9}\frac{m}{s}##
##m=1.2⋅10^4kg##
##θ=15°##
##K=\frac{1}{2}mv^2##
##-K=-Fy=-(mgsinθ)(y)##
##y=Lsinθ##
The Attempt at a Solution
First, I figured that the gravitational work must equal to the work produced by the truck and its velocity. And gravity does work only in the vertical direction, which is why I'm expressing gravitational work in terms of the force and the vertical distance ##y##. This is what I did, and it isn't giving me the correct answer.
##K=\frac{1}{2}mv^2##
##-K=-Fy=-(mgsinθ)(y)##
##K=Fy=(mgsinθ)(y)=(mgsinθ)(Lsinθ)##
##(mgsinθ)(Lsinθ)=(\frac{1}{2}mv^2)##
##(L)(g)(sin^2θ)=(\frac{1}{2}v^2)##
##L=(\frac{1}{2g})(v^2)(csc^2θ)##
Evaluating this gives me: ##L=993.2m##, which is not the answer from the back of the book. I feel like there's something I'm not understanding. Can someone help me? Thanks.