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3 December, 10:11

Two blocks A and B with mA = 2.2 kg and mB = 0.84 kg are connected by a string of negligible mass. They rest on a frictionless horizontal surface. You pull on block A with a horizontal force of 5.7 N. (a) Find the magnitude of the acceleration (in m/s2) of the blocks. m/s2 (b) Determine the tension (in N) in the string connecting the two blocks. N (c) How will the tension in the string be affected if mA is decreased? increase decrease remain the same

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  1. 3 December, 12:14
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    (a) a = 1.875 m/s²

    (b) T = 1.575 N

    (c) T increase

    Explanation:

    Newton's second law:

    ∑F = m*a Formula (1)

    ∑F : algebraic sum of the forces in Newton (N)

    m : mass s (kg)

    a : acceleration (m/s²)

    We define the x-axis in the direction parallel to the movement of the blocks on the horizontal surface and the y-axis in the direction perpendicular to it.

    Forces acting on the block A

    WA: Weight of of the A block : In vertical direction downaward (-y)

    NA : Normal force of the A block : In vertical direction upaward (+y)

    F = 5.7 N In in the direction parallel to the movement of the blocks (+x)

    T : tension of the string: In in the direction (-x)

    Forces acting on the block B

    WB: Weight of the B block: In vertical direction downaward (-y)

    NB : Normal force of the B block : In vertical direction upaward (+y)

    T : tension of the string: In in the direction (+x)

    Data

    mA = 2.2 kg

    mB = 0.84 kg

    (a) Magnitude of the acceleration of the blocks

    Newton's second law to B block:

    ∑Fx = m*a

    T = (0.84) * a Equation (1)

    Newton's second law to A block:

    ∑Fx = m*a

    5.7 - T = (2.2) * a We replace T of the Equation (1)

    5.7 - (0.84) * a = (2.2) * a

    5.7 = (2.2) * a + (0.84) * a Equation (2)

    5.7 = (3.04) * a

    a = 5.7 / (3.04)

    a = 1.875 m/s²

    (b) Tension (in N) in the string connecting the two blocks

    We replace data in the Equation (1)

    T = (0.84) * a

    T = (0.84) * (1.875)

    T = 1.575 N

    (c) How will the tension in the string be affected if mA is decreased?

    We observed Equation (2)

    5.7 = (2.2) * a + (0.84) * a

    5.7 = (mA) * a + (0.84) * a

    5.7 = a * (mA + 0.84)

    if mA decrease, then, the acceleration increase and T increase
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