A 2kg steel ball, initially at rest, rolls down a ramp. It experiences an acceleration of 1.2m/s^2. If it rolls a distance of 2.8m, how fast is it travelling? What is the angle of elevation of the ramp?
Ok, so i got the 1st part and i got a final velocity of 2.6m/s but i have no idea how to find theta(angle). Can anyone help me out on this?
2006-12-30 10:38:33 · 5 answers · asked by My Answers are always right. 2 in Science & Mathematics ➔ Physics
well im still kind of lost, im sure that they are both related but could you use something like 1.2sin9.8 to find theta? Or would that be wrong?
2006-12-30 11:00:57 · update #1
If you were to drop an object vertically down near the surface of the earth, the acceleration due to gravity is 9.81 m s^(-2). Acceleration down an inclined plane is still caused by gravity, but the acceleration is weaker because there is a normal force imposed by the inclined plane on the object. Do you know how to relate the acceleration down a plane inclined at an angle theta to the vertical acceleration of gravity? (hint: it has something to do with sine theta) Once you figure that out, finding theta shouldn't be too difficult, since you are given 1.2 m s^(-2), and you know the vertical acceleration due to gravity is 9.81 m s^(-2). Hope this helps without spelling out the answer.
2006-12-30 10:46:36 · answer #1 · answered by Telodrift 2 · 0⤊ 0⤋
I hope you have had exercises finding the resultant of 2 vectors. And that you can do the reverse: take a single vector and break it into 2 perpendicular components.
That's what you need to do. The ball accelerates down the ramp because there is a component of g that points down the slope. This component is the acceleration 1.2 m/s^2. The other component is normal to the slope (perpendicular to the vector pointing down the ramp). You need to draw g and these 2 components. g is the resultant, and if you use the parallelogram method, g is the hypotenuse of a triangle. You know g and you know one of the other sides. From here, use some trigonometry.
2006-12-30 11:14:57 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋
set up a coordinate system with the x-axis pointing down
the ramp and the y-axis perpendicular to the ramp
think about the forces acting on the ball. since the ball is
accelerating, the forces must not add up to zero, i.e. there
is a net force. since the ball is accelerating *down* the ramp,
this must also be the direction of the net force (that's why
i suggested letting the x-axis point *down* the ramp)
there are 2 forces: the weight = mg, which acts vertically
down (it always does) and the contact force i.e. due to the
ramp, which acts in the y-direction (often called the normal
force).
add up the components of the forces in the x-direction (down
the ramp). the contact force has no x-component, so
the acceleration down the ramp is entirely due to the
x-component of the weight. therefore, newton's law says
x-component of the weight = mass * acceleration
or
mg sin(theta) = ma
where theta is the angle of the ramp.
cancel m and solve for theta !
2006-12-30 11:16:27 · answer #3 · answered by john_doe 1 · 0⤊ 0⤋
What would the acceleration be if the ramp wasn't there?
What does that tell you about the angle between the ramp and the accelerating force?
2006-12-30 10:58:37 · answer #4 · answered by virtualguy92107 7 · 0⤊ 0⤋
Cutting through all the physics bs here, the angle you're looking for has a sin of a/g = 1.2/9.8 = .1224 â α = 7.033°
2006-12-30 11:47:09 · answer #5 · answered by Steve 7 · 0⤊ 0⤋