(II) How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.3 m across a rough floor without acceleration, if the effective coefficient of friction was 0.50?

Respuesta :

Answer:

2324 J

Explanation:

The formula for work is:

[tex]W=F*d[/tex]

where [tex]F[/tex] is the force applied, and [tex]d[/tex] is the distance moved, in this case [tex]d=10.3m[/tex]

and we need to find [tex]F.[/tex]

Since the crate is not moving up or down, we conclude that the normal force must be equal to the weight of the object:

[tex]N=w[/tex]

where [tex]N[/tex] is the normal force and [tex]w[/tex] is the weight, which is: [tex]w=mg[/tex], where g is the gravitational acceleration [tex]g=9,81m/s^2[/tex] and [tex]m[/tex] is the mass [tex]m=46kg[/tex].

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Thus the normal force is:

[tex]N=mg[/tex]

Now, the force due to the friction is defined as:

[tex]f=\mu N=\mu mg[/tex]

where [tex]\mu[/tex] is the coefficient of friction, [tex]\mu =0.5[/tex]

So, for the crate to move, the force applied must be equal to the frictional force:

[tex]F=f\\F=\mu mg[/tex]

And now that we know the force we can calculate the work:

[tex]W=F*d\\W=\mu mg*d[/tex]

substituting known values:

[tex]W=(0.5)(46kg)(9.81m/s^2)(10.3)\\W=2324J[/tex]

Answer:

2321.62 J

Explanation:

From the question,

F = μR..................... Equation 1

Where, F = Frictional force, μ = coefficient of static friction, R = normal reaction

R = mg.................................. Equation 2

Where, m = mass of of the crate, g = acceleration due to gravity

Substitute equation 2 into equation 1

F = mgμ........................ Equation 3

Given: m = 46 kg, g = 9.8 m/s², μ = 0.5

F = 46(9.8)(0.5)

F = 225.4 N

Also,

W = F×d.................. Equation 3

Work = Work done, F = Force, d = distance

Given: F = 225.4 N, d = 10.3 m

W = 225.4×10.3

W = 2321.62 J