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Physics (Algebra-based)

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A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track.It leaves the track horizontally, flies through the air, and subsequently strikes the ground. gravity= 9.81 m/s^2.
height1=4.1m---- is initially dropped from here
height2=2.4m---- part where curve flattens out
mass of block=0.546kg
What is the speed v of the block when it
leaves the track (in m/s)?
What is the horizontal distance x the block travels in the air (in m)?
What is the speed of the block when it hits the ground (in m/s)?
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  • Physics (Algebra-based) - ,

    PE → KE1
    KE1 → Work(fr) +KE2 =>
    PE → Work(fr) +KE2
    If the initial height is ‘h₁’,
    the height of the horizontal section is ‘h₂’,
    the length of the rough section is ‘s’, then
    mg (h₁-h₂)=μmgs +mv²/2,
    v=sqrt{ 2g[(h₁-h₂)-μs]}.
    ---
    h₂=gt²/2,
    t=sqrt(2h₂/g).
    x=vt,
    v(y)=gt.
    V(near the ground)=sqrt{v²+v(y)²}

  • Physics (Algebra-based) - ,

    Thank you!

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