Question

how much work did the movers do (horizontally) pushing a 160-kg crate 10.3 m across a rough floor without acceleration, if the effective coefficient of friction was 0.50?

Answer 1

Answer:

The work done is 8075.2 J.

Explanation:

Given that,

Mass of crate = 160 kg

Distance = 10.3 m

coefficient of friction = 0.50

We need to calculate the normal force

Using balance equation

$F_{y}=ma_{y}$

$F_{N}-F_{g}=ma_{y}$

Here, acceleration a = 0

$F_{N}=F_{g}$

$F_{N}=mg$

Put the value into the formula

$F_{N}=160\times9.8$

$F_{N}=1568\ N$

We need to calculate the frictional force

Using balance equation

$F_{x}=ma_{x}$

$F_{p}-f_{kx}=0$

$F_{p}=\mu mg$

$F_{p}=0.50\times160\times9.8$

$F_{p}=784\ N$

We need to calculate the work done

Using formula of work done

$W=F_{p}\ d\cos\theta$

$W=784\times10.3\cos0$

$W=8075.2\ J$

Hence, The work done is 8075.2 J.

Answer 2

We are given with a 160 kg crate in which it is moved by 10.3 meters and the coefficient of friction of the floor is equal to 0.50. The formula is expressed as W = F u d. substituting, W = 160 kg * 9.8 m/s2 * 0.5* 10.3 m equal to 8075.2 joules.