- #1
ericcy
- 19
- 1
- Homework Statement
- A 1975kg car is parked at the top of a steep 42m long hill inclined at an angle of 15 degrees. If the car starts rolling down the hill, how fast will it be going when it reaches the bottom of the hill (neglect friction).
- Relevant Equations
- Ei=Ef, Ek=1/2mv^2, Ep=mgh
I determined 42m to be the hypotenuse so I used sine law to find the height of the incline, 10.87m. I used this height in the equation Ei=Ef, since they should be equal.
Ei=Ef
mgh=1/2mv^2 (at the start there is no kinetic energy, at rest. at the end there is only kinetic, no potential)
1975(9.81)(10.87)=1/2(1975)v^2
210603.532/987.5=987.5v^2/987.5
(square root)213.269=(square root)v^2
V=14.603
Answer in the back of the book is 0.4m/s, for whatever reason. All responses are appreciated, thanks.
Ei=Ef
mgh=1/2mv^2 (at the start there is no kinetic energy, at rest. at the end there is only kinetic, no potential)
1975(9.81)(10.87)=1/2(1975)v^2
210603.532/987.5=987.5v^2/987.5
(square root)213.269=(square root)v^2
V=14.603
Answer in the back of the book is 0.4m/s, for whatever reason. All responses are appreciated, thanks.