An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has four times th

Question

An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has four times the mass. Its acceleration will be 2 A. 1 m/s . 2 B. 2 m/s . 2 2 Which of the following pairs of acceleration of the blocks and tension in the rope is correct? A. Acceleration = 3.19 m/s2 and Tension = 702.5 N. B. Acceleration = 4.05 m/s2 and Tension = 297.5 N. C. Acceleration = 4.05 m/s2 and Tension = 702.5 N. D. Acceleration = 3.19 m/s2 and Tension = 297.5 N. C. 4 m/s D. 8 m/s E. None of the above.

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Sigridomena 6 months 2021-07-15T01:44:22+00:00 1 Answers 34 views 0

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    2021-07-15T01:45:49+00:00

    Answer:

    B. 2 m/s

    B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

    Explanation:

    A force is applied on a mass m whose acceleration is 4 m/s

    Force = mass × acceleration

    a = F/m = 4 m/s

    4 m/s = F/m

    F = 4 m/s (m)

    If  Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

    2F/4 m = F/ 2m

    4m/s (m) / 2m = 2 m/s

    a = 2 m/s

    2.

    Given that

    mass m_1 = 30 kg

    mass m_2 = 50 kg

    \mu = 0.1

    From the question; we can arrive at two cases;

    That :

    m_{2} a _ \ {net} }= m_2g - T   —– equation (1)

    m_{1} a _ \ {net} }=  T - mg sin \theta  - F —- equation (2)

    50 a = 50 g – T

    30 a = T – 30 g sin 30 – 4 × 30 g cos 30

    By summation

    80 a =[ 50  - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]g

    80 a = 32. 4 × 10 m/s ²  (using g as 10m/s²)

    80 a = 324 m/s ²

    a = 324/80

    a = 4.05 m/s²

    From equation , replace a with 4.05

    50 × 4.05 = 50 × 10 – T

    T = 500 -202.5

    T =297.5 N

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