Question

(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?

Answer 1

Answer:

2324 J

Explanation:

The formula for work is:

$W=F*d$

where $F$ is the force applied, and $d$ is the distance moved, in this case $d=10.3m$

and we need to find $F.$

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

$N=w$

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

---------

Thus the normal force is:

$N=mg$

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

$f=\mu N=\mu mg$

where $\mu$ is the coefficient of friction, $\mu =0.5$

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

$F=f\\F=\mu mg$

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

$W=F*d\\W=\mu mg*d$

substituting known values:

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

Answer 2

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