Physics problem, help me please?

Comments

• gp4rts

  May 2021

  You can also get the distance from energy consideration:

  Friction force = ¨µ*m*g*cosθ

  Sliding force = m*g*sinθ

  At constant velocity these are equal so µ*m*g*cosθ = m*g*sinθ and

  µ = tanθ

  The initial energy is 0.5*m*vB²

  this must equal the energy used to reach the stopping point

  m*g*h + Fr*L

  h = L*sinθ; Fr = µ*m*g*cosθ = m*g*sinθ

  0.5*m*vB² = m*g*L*sinθ + m*g*sinθ*L = 2*m*g*sinθ*L

  L = 0.25*vB²/(g*sinθ)

  After stop, the sliding force is m*g*sinθ. For the block to slide down, this must exceed the static friction force, Fs. Fs = µs*m*g*cosθ

  m*g*sinθ > µs*m*g*cosθ

  µs < tanθ < µ; This is almost never the case (static friction is more than sliding friction), so the block does not slide down again.
• kuiperbelt2003

  May 2021

  if the block slides down the plane with constant velocity, then the forces acting on it are equal (since there is no acceleration)

  the forces are the component of gravity down the plane and friction up the plane

  Newton's second law for this is

  mg sin(A)-u mg cos(A)=0 where u is the coefficient of friction between the block and slide

  therefore we have g(sin(A)-u cos(A))=0 or

  tan(A)=u

  for the block sliding up the plane, both gravity and friction act down the plane, so Newton's second law becomes

  -mg sin(A)-u mgcos(A) = ma

  -g(sin(A) + u cos(A))=a

  since u=tan(A) = sin(A)/cos(A), we have

  a=-g(sin(A)+sin(A))=-2gsin(A)

  so we see in this case that there is a net negative acceleration in the case of sliding up the plane; we use

  vf^2=v0^2 + 2ad where vf=final speed = 0 at top

  v0=initial speed = B

  a=acceleration = -2gsin(A)

  d=distance, so

  0=B^2 +2(-2gsin(A))d or

  d=B^2/4gsinA

  is the distance the block slides up the plane

  b) the u calculated in part a) is the coefficient of kinetic friction; to determine whether the block will slide from rest requires a knowledge of the coefficient of static friction; and the only statement we can make is:

  us > uk

  so applying Newton's second law to the stationary block at the top of its motion, we have:

  mg sinA - us cos A =ma

  a=g(sinA-us cosA)

  but us>uk=sinA/cosA so

  a<g(sinA-sinA) so a is less than zero meaning the block will not slide down once it reaches the top of its motion on the plane