The discussion revolves around calculating the velocity of a box as it leaves a frictionless ramp. Participants are exploring the effects of gravity and the forces acting on the box during its motion along the ramp.
The conversation is active, with participants questioning assumptions about the forces involved and the calculations needed. Some guidance has been offered regarding the relevance of the initial push and the need to consider forces as vectors, but no consensus has been reached on the calculations themselves.
There is uncertainty regarding the mass of the box, which is not provided, and participants are considering it as an arbitrary variable. The discussion also involves clarifying the definitions of terms like "fixed" and "frictionless" in the context of the ramp.
Is gravity the only force acting on the box? In which direction is it accelerating?goonking said:a is gravity so it is -9.8, correct?
What is the distance the box travels?goonking said:but what do i put in as x? 5, which is the length of the ramp, or 4, the base, or 3 the height?
I'm not sure what you mean by fixed, but it is friction less.Nathanael said:The ramp is fixed and frictionless, right?Is gravity the only force acting on the box? In which direction is it accelerating?What is the distance the box travels?
I mean it can't move.goonking said:I'm not sure what you mean by fixed, but it is friction less.
Nathanael said:I mean it can't move.
What direction, what distance, and with what acceleration does the block move?
It slows down the whole way. But at what rate?goonking said:the ramp is fixed.
the box goes up the ramp.
the distance it travels before flying off the ramp is 5 meters.
and I have no idea how to calculate the acceleration. I assume it slows down near the top since it is going against gravity.
Since it's frictionless, the only forces are gravity and normal force. correct? also the force from the initial pushNathanael said:It slows down the whole way. But at what rate?
Have you tried drawing a free body diagram? What are all the forces acting on the block?
Right, but the force from the initial push is irrelevant. We can "start our clocks" immediately after that force stops; when the block is moving at 12 m/sgoonking said:Since it's frictionless, the only forces are gravity and normal force. correct? also the force from the initial push
direction of normal force is perpendicular to the surface.Nathanael said:Right, but the force from the initial push is irrelevant. We can "start our clocks" immediately after that force stops; when the block is moving at 12 m/s
What is the size and direction of the normal force? And what is the net force that results from adding gravity and the normal force together?
I edited out my posit made no sense, to have Fg, we need the mass of the object, and that isn't given.Nathanael said:Yes but it's time to get into equations: What is the size of the normal force? More importantly, what is the size of the net force?
but it should be cos theta = Fg⊥ to surface divided by Fg. correct?Nathanael said:Yes but it's time to get into equations: What is the size of the normal force? More importantly, what is the size of the net force?
Maybe all hope is not lost, just continue as normal. The mass is hereby an arbitrary "m"goonking said:to have Fg, we need the mass of the object, and that isn't given.
Right, so that gives you |F⊥| = |Fg|cosθgoonking said:but it should be cos theta = Fg⊥ to surface divided by Fg. correct?
|F⊥| = |Fg|cosθ= 7.84Nathanael said:Maybe all hope is not lost, just continue as normal. The mass is hereby an arbitrary "m"Right, so that gives you |F⊥| = |Fg|cosθ
So what is the sum of the two vectors F⊥+Fg ?
No, these are vectors. They have directions. You must add them component by component. Consider the component of gravity perpendicular to the ramp... How does it combine with the normal force? What net force do they produce in the direction perpendicular to the ramp? Then what is left of gravity?goonking said:|F⊥| = |Fg|cosθ= 7.84
Fg = 9.8
so adding them is 9.8 + 7.84 = 17.64
is that correct?