Hello Juggalomike Quote:

Originally Posted by

**Juggalomike** This question is killing me, i believe it is solved with triangles(which is why im posting it in the geometry forum) because every method i have tried has failed.

The speed of a 2.5 kg block sliding directly down a frictionless inclined plane is found to be 1.70 m/s. At 1.51 seconds later it has a velocity of 5.72 m/s. Calculate the angle of the plane with respect to the horizontal (in degrees).

I was thinking that the vertical axis would be 9.8 for gravity so i could simply do ((5.72-1.70)/1.51)/9.8 to get the % of 9.8 that the acceleration is, then multiply 90 degrees by that, the angle seems as if it would work but im getting the wrong answer, anyone else have any idea?

If the angle the plane makes with the horizontal is $\displaystyle \alpha$, then, when we resolve the forces down the plane, the equation of motion is:

$\displaystyle mg\sin\alpha = ma$

where $\displaystyle m$ is the mass of the body, and $\displaystyle a$ is its acceleration down the plane.

$\displaystyle \Rightarrow \sin\alpha = \frac{a}{g}$$\displaystyle =\frac{5.72-1.70}{1.51\times9.8}$

$\displaystyle \Rightarrow \alpha=15.75^o$

Grandad