6. 3. A 7.6 kg object is pulled 6.0 m at a constant velocity of 5.0 m/s along a horizontal surface by a force of 2.

Question

6.
3. A 7.6 kg object is pulled 6.0 m at a constant
velocity of 5.0 m/s along a horizontal surface by
a force of 2.0 N. What is the work done on the
object to overcome friction?

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Calantha 3 years 2021-08-30T14:15:09+00:00 1 Answers 5 views 0

Answers ( )

  1. linhdan
    0
    2021-08-30T14:16:28+00:00
    Reply

    Answer:

    12 J

    Explanation:

    From the question given above, the following data were obtained:

    Mass (m) = 7.6 kg

    Distance (d) = 6 m

    Velocity (v) = 5 m/s

    Force (F) = 2 N

    Workdone (Wd) =.?

    Workdone can be defined as the product of force and distance moved in the direction of the force. Mathematically, it is expressed as:

    Workdone = Force × distance

    Wd = F × d

    With the above formula, we can obtain the workdone as follow:

    Distance (d) = 6 m

    Force (F) = 2 N

    Workdone (Wd) =.?

    Wd = F × d

    Wd = 2 × 6

    Wd = 12 J

    Thus, the workdone is 12 J

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )