A block of mass m=0.280kg is released from point a, at heighth=0.100m on a U-sha

Question

A block of mass m=0.280kg is released from point a, at heighth=0.100m on a U-shaped track. It slides down the track and up topoint c, where it is momentarily at rest, before sliding back downand back up the other side. The height at point c is h=0.030m. Thedistance along the track from point a to point b is 0.157m andpoint c is 0.078m beyond that (along the track). There isfriction.

The question is

A. Calculate the work done by friction on the block betweenpoints a and c.

B. Calculate the average frictional force.

Explanation / Answer

The mass of the block is m = 0.280 kg The height of the U-shaped track is h = 0.100 m The height at point c is h = 0.030 m The distance along the track from point a to point b is 0.157mand point c is 0.078m beyond that (along the track). A.When the block is released from point a and reaches the baseof the U-shaped track,then the force of friction is Fy= n + (-w) = fk1 - w or fk1= w The force of friction on the block when the block movesthrough the base of the U-shaped track is Fx= F + (-fk2) = 0 or F = fk2 The force of friction on the block when the block moves upthrough the base of the U-shaped track is Fy= n + (-fk3) = 0 or n = fk3 Therefore,the force acting on the block due to frictionis Fk= fk1 + fk2 +fk3 Here,fk1= fk2= fk3=k* n or Fk= k* n + k*n + k* n or Fk= 3k* n Here,n = w or Fk= 3k* w =3k* m * g Here,k= 1,m = 0.280 kg and g = 9.8m/s2 or Fk= 1 * 0.280 * 9.8 = 2.744 N Therefore,the workdone by friction on the block between pointsa and c is W = Fk * S Here,S = 0.157 + 0.078 = 0.235 m or W = 2.744 * 0.235 or W = 0.64 J B.The average frictional force is Fk' = (Fk/3) or Fk' = (2.744/3) = 0.91 N C.Let the velocity at point b be v,therefore,we get m * g * h = (1/2)mv2 or v2= 2 * g * h or v = (2 * g * h)1/2 Substituting the values in the above equation,we get v = (2 * 9.8 * 0.100)1/2 or v = 1.4 m/s or v = 1.4 m/s

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