Two blocks with masses 4.00 kg and 8.00 kg are connected by string and slide down a 30.0 degree inclined plane. The coefficient of kinetic friction between the 4.00-kg block and the plane is 0.25; that between the 8.00-kg block and the plane is 0.35.a) Calculate the acceleration of each blockb) Calculate the tension in the stringc) What happens if the positions of the block are reversed, so that the 4.00-kg block is above the 8.00-kg block?

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hajavaF · Accepted Answer

First note that you must establish a reference frame (i.n. xy plane). For convenience, it is generally assumed that thex-axis is parallel to the slope of the surface and the y-axis is perpendicular to the slope. Furthermore, the positive x-direction is down slope.  Given the above assumptions, now determine the sum of the forces for each block.  m1=4kg mass of block 1  m2=8kg mass of block 2  u1=0.25 coefficient of friction for block 1  u2=0.35 coefficient of friction for block 2  α=30 deg → slope of ramp  g = acceleration due to gravity = 9.8m/s  sum of forces on block 1:  Note that the coefficient of friction is applied to the force normal to the surface  ∑F1=m1⋅gsin⁡(α)−u1⋅m1⋅g⋅cos⁡(α)=11.1129 newtons  where m1⋅g⋅sin⁡(α) is the force in the positive x-directionon block 1 due to its weight and m1⋅g⋅cos⁡(α) is the normal force, perpendicular to the slope, due to acceleration of gravity.  Given the sum of forces for F1, its

Jeffrey Jordon · Accepted Answer

n1​=m1​gcos⁡30  fk1=μk1m1gcos⁡30  m1a1=m1gsin⁡30−μk1m1gcos⁡30  a1=9.8(sin⁡30−0.2cos⁡30)=3.20ms2  n2=m2gcos⁡30  fk2=μk2m2gcos⁡30  m2a2=m2gsin⁡30−μk2m2gcos⁡30  a2=9.8(sin⁡30−0.4cos⁡30)=1.51ms2  b)  m1a1=m1gsin⁡30−μk1m1gcos⁡30−T  m2a2=m2gsin⁡30−μk2m2gcos⁡30+T  T=4⋅8(3.2−1.51)4+8=4.51N

Two blocks with masses 4.00 kg and 8.00 kg are connected by string and slide down a 30.0 degree inclined plane.

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