- #1

- 37

- 0

Okay, so the question I set myself is; If a box of mass 10 KG is put at the top of a ramp at 60 degrees and with a hypotenuse of length 6m, what will be the velocity of the box at the bottom of the ramp? (I included friction into the problem. I used a hypothetical kinetic coefficient of friction of 0.5).

So the known variables are: Ramp angle = 60 degrees.

Box mass = 10 KG

Kinetic Coefficient of friction = 0.5

Length of ramp (hypotenuse, not adjacent) = 6m

So here's what I did...

1) I found the force of gravity acting along the ramp. I found the force of gravity by doing the following.. 9.81*COS(90-60) = 8.5 m/s^2. Since F=ma, I found the component of gravity acting along the ramp to be 85N.

2) Next I found the same for friction. Since friction is the normal force times by the coefficient of friction, i did 85 (normal force) times by 0.5 (coefficient). This comes to 42.5.

3) To find the net force, I did the force of gravity (85N) minus the force of friction (42.5N). This comes to a total of 42.5N of force down the ramp.

4) Since F=ma, I found the acceleration to be 4.25 m/s^2.

5) With the formula v^2=2as, I found that 2*4.25*6=51. Finally, I found the square root of 51 to be 7.14.

This shows that the velocity at the bottom of the ramp is 7.14 m/s.

B.T.W, can anyone confirm whether this type of question is likely to come up in my GCSE next year? I want to get a solid idea on all of the topics while I can.