Ask Question
27 August, 16:57

A 50.0-kg box is being pushed along a horizontal surface by a force of 250 n directed 28.0o below the horizontal. the coefficient of kinetic friction between the box and the surface is 0.300. what is the acceleration of the box?

+1
Answers (1)
  1. 27 August, 17:09
    0
    Refer to the figure shown below.

    Tha applied force of F = 250 N has

    (i) a horizontal component of Fx = (250 N) cos (28°) = 220.737 N,

    (ii) an upward vertical component of Fy = (250 N) sin (28°) = 117.368 N

    The weight of the box is

    W = (50 kg) * (9.8 m/s²) = 490 N

    The normal reaction on the box is

    N = W - Fx = 490 - 117.368 = 372.632 N

    The resistive frictional force on the box is

    μN = 0.3*372.632 = 111.790 N

    The horizontal driving force on the box is

    Fx - μN = 220.737 - 111.790 = 108.947 N

    If the acceleration of the box is a m/s², then

    (50 kg) * (a m/s²) = (108.947 N)

    a = 108.947/50 = 2.179 m/s²

    Answer:

    The acceleration of the box is 2.18 m/s² (nearest hundredth)
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A 50.0-kg box is being pushed along a horizontal surface by a force of 250 n directed 28.0o below the horizontal. the coefficient of ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers