Forces on inclined plane question

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  • #1
Jake14
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1. An object of mass m = 1 kg is initially placed on a frictionless plane, inclined by an angle θ = 30o as indicated below. How long does it take the mass to move the distance of 10 m down the incline?


2. I had a question if how I solved this is correct. I got the right answer and I wanted to make sure this was not luck. I know that we can use kinematic equations here but I tried something else...here is what I did.


3. So we know that v=(d/t) we know d = 10m and we are solving for t, so we just need v. I know the F = ma
and I know the mass and the force on the block. The force is the x-component of the gravitational force, so Fx = mgsinθ so we can write
mgsinθ = ma and we know that a = (dv/dt) so mgsinθ = m*(dv/dt) so let's divide by m and move the dt to the left side so we get
dv = gsinθ dt. Let's integrate this so v = gsinθ*t plug this in for v in the v = (d/t) equation so
t^2*g*sinθ = d
you can now solve for t and I got an answer close to the multiple choice answer. Is this correct how I solved for velocity?

-jake
 
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  • #2
haruspex
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V=d/t only works for constant velocity, or if what you are after is average velocity.
The answer you got from your last equation should be wrong, only about 2/3 of the correct answer.
Are you familiar with the SUVAT equations? These apply for constant acceleration.
 
  • #3
24forChromium
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1. An object of mass m = 1 kg is initially placed on a frictionless plane, inclined by an angle θ = 30o as indicated below. How long does it take the mass to move the distance of 10 m down the incline?
I am not sure what is so difficult about this question, just find the component of the gravitational force that does work by trigonometry and use this equation to find out the time taken:

a = F_ge / 1kg (where F_ge is the effective acceleration)
(a*t^2)/2 = 10m where t is the total time required
 

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