• February 4th 2007, 09:21 AM
pablohacker
Hello all, I am a little bit stuck on the following question and the types that are like it, so could you please help me. :confused:

The question is in full is

"A boy is sliding down a slope, inclined at 25 degrees to the horizontal. The resistance to this motion amount to 15N. Find the mass of this boy when his acceleration is 2.9ms."

• February 4th 2007, 11:52 AM
topsquark
Quote:

Originally Posted by pablohacker
Hello all, I am a little bit stuck on the following question and the types that are like it, so could you please help me. :confused:

The question is in full is

"A boy is sliding down a slope, inclined at 25 degrees to the horizontal. The resistance to this motion amount to 15N. Find the mass of this boy when his acceleration is 2.9ms."

Hmmm...The way the problem is worded makes it sound like the boy's mass is changing! :eek:

Anyway, start with a Free-Body Diagram. I've got a weight force (w) acting straight downward, a normal force (N) directly out of the slope, and a resistance force (R) acting up the slope. I'm going to define a coordinate system such that +x is down the slope and +y is in the direction of the normal force.

Since the question deals with an acceleration in the x direction I'm going to do Newton's second in the x direction:
$\sum F_x = w_x - 15 = ma$
where $w_x$ is the x component of the weight force: $w_x = mg sin(\theta)$ where $\theta$ is the angle of the incline. (Make sure you understand why this is the sine function and not cosine.)

So:
$mgsin(\theta) - 15 = ma$

You know a and, obviously, g, so we can solve for m:
$m = \frac{15}{gsin(\theta) - a}$

I get $m \approx 12.0806 \, kg$

-Dan
• February 5th 2007, 01:45 AM
pablohacker