Consider a very low (zero) friction 5.0 kg skateboard on a ramp at an angle of 15 degrees to the horizontal.

1. What would be the net force that causes the acceleration when the skateboard is allowed to move?
2. What would the skateboard's acceleration be down the plane?
3. Now consider the same no-friction skateboard on the same 15-degree ramp. If a 45 kg teenager jumps on, what would be her acceleration down the ramp?

You need to use trigonometry to work this one out. The components of the force are needed. it is a vector, so work out the unbalanced force using trig. you know that the force pulling the 5kg skateboard down is going to be 9.8 * 5 = 49N. As you have the angle it is on, and 49 is the vertical component, you can use Sin(X) = Op/Hyp where x= 15 degress, the opposite is the vertical component (49N) and the hyp is the unbalanced force - which is 79.35. I think. Somebody plz double check me on that one though :P

Answered by Gwenuin | 2024-06-10

The net force acting on the 5.0 kg skateboard on a 15-degree ramp is approximately 12.66 N, leading to an acceleration of about 2.53 m/s² down the incline. When a 45 kg teenager jumps on, the total mass becomes 50.0 kg, but the acceleration down the ramp remains the same at approximately 2.53 m/s² due to the unchanged incline angle. This shows that, on an incline without friction, the acceleration is independent of mass.
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Answered by Gwenuin | 2024-10-31