Anonymous
Anonymous asked in Science & MathematicsPhysics · 5 months ago

physics help?

(a) 2.0-kg mass is suspended by a string from the ceiling of an elevator that is traveling downward at a constant speed of 1.0 m/s. What is the tension in the string?

(b) A 2.0-kg mass is suspended by a string from the ceiling of an elevator that is accelerating downward at 1.4 m/s2. What is the tension in the string?

(c) A 2.0-kg mass is resting on a scale in an elevator that is accelerating upward 1.4 m/s2. What is the reading on the scale

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  • NCS
    Lv 7
    5 months ago

    tension T = m*(g ± a)

    where a is the kinematic acceleration

    ... use "+" if accelerating against gravity

    ... and "-" if accelerating with gravity

    given m = 2.0 kg

    and g = 9.8 m/s²

    (a) a = 0, so T = 20 N

    (b) a = -1.4 m/s², so T = 17 N

    (c) a = +1.4 m/s², so T = 22 N

    All answers to two significant digits, consistent with the data.

    Hope this helps!

  • oubaas
    Lv 7
    5 months ago

    (a)

    2.0-kg mass is suspended by a string from the ceiling of an elevator that is traveling downward at a constant speed of 1.0 m/s. What is the tension in the string?

    T = m*g = 2*9.807 = 19.61 N (20 N with 2 significant figures)

    (b)

    A 2.0-kg mass is suspended by a string from the ceiling of an elevator that is accelerating downward at 1.4 m/s2. What is the tension in the string?

    T = m*(g-a) = 2*(9.807-1.4)  = 16.81 N (17 N with 2 significant figures)

    (c)

    A 2.0-kg mass is resting on a scale in an elevator that is accelerating upward 1.4 m/s2. What is the reading on the scale

    T = m*(g+a) = 2*(9.807+1.4) = 22.41 N (22 N with 2 significant figures)

  • 5 months ago

    By Tension (T) = F(gravity) +/- F(due to lift acceleration) {-ve for downward & + ve for upward motion of the lift}

    =>T = m x (g +/-a) --------------------(x)

    (a) As v = constant=>a = 0

    =>T = mg = 2 x 9.8 = 19.6 N

    (b) T = m x (g - a) = 2 x (9.8 - 1.4) = 16.8 N

    (c) T = m x (g + a) = 2 x (9.8 + 1.4) = 22.4 N

  • 5 months ago

    a) m = 2.0-kg mass is suspended by a string from the ceiling of an elevator that is traveling downward at a constant speed of V = 1.0 m/s. What is the tension in the string?

    As A = 0 there is no force other than the mass weight W = mg = 2*9.8 = 19.6 N for the tension.  ANS.(b) A 2.0-kg mass is suspended by a string from the ceiling of an elevator that is accelerating downward at 1.4 m/s2. What is the tension in the string?At A = 1.4 m/s^2, that weight W becomes net weight w = W - Fy = M(g - A) = 2*(9.8 - 1.4) = 16.8 N  tension.  ANS.(c) A 2.0-kg mass is resting on a scale in an elevator that is accelerating upward 1.4 m/s2. What is the reading on the scale

    And as I hope you've guessed by now, the net weight w = W + Fy = 2*(9.8 + 1.4) = 22.4 N tension.  ANS.

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