Two blocks are sliding to the right across a horizontal surface, as the drawing

Question

Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In Case A the mass of each block is 3.1 kg. In Case B the mass of block 1 (the block behind) is 6.2 kg, and the mass of block 2 is 3.1 kg. No frictional force acts on block 1 in either Case A or Case B. However, a kinetic frictional force of 6.1 N does act on block 2 in both cases and opposes the motion. For both Case A and Case B determine the following. (a) the magnitude of the forces with which the blocks push against each other Case A: P = Case B: P = (b) the magnitude of the acceleration of the blocks Case A: a = Case B: a =

Explanation / Answer

Let, force with which the blocks push against each other = FAB

Equation for block 1,

-FAB = m1*(-a)

a = FAB / m1 ......(1)

Equation for block 2,

FAB - f = m2*(-a)

Put the value of a from eq(1),

FAB - f = m2 * (-FAB / m1)

FAB = m1*f / (m1 + m2)

For case A,

FAB = 3.1*6.1 / (3.1 + 3.1)

FAB = 3.05 N

For case B,

FAB =  6.2*6.1 / (6.2 + 3.1)

FAB = 4.06 N

(b)

For case A,

magnitude of the acceleration of the blocks,

a = FAB / m1 = 3.05 / 3.1

a = 0.98 m/s^2

For case B,

a = FAB / m1 = 4.06 / 6.2

a = 0.655 m/s^2

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