Question

How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.4 m across a rough floor without acceleration, if the effective coefficient of friction was 0.60? Express your answer using two significant figures.

Answer 1

Answer:

The total work done by the mover is 2.81 kJ.

Explanation:

Given the 46 kg crate is displaced by 10.4 meters.

And the acceleration is zero. Also, $\mu_k=0.60$

let $P$ is applied force, $F_N$ is the net force, $m$ is the mass and $g=9.81\ m/s^2$

and $\mu_k=0.60$

$F_N=P- \mu_k\times mg$

As the acceleration is zero, the net force will also be zero.

$0=P- \mu_k\times mg\\P=\mu_k\times mg$.

$P=0.6\times 46\times 9.81=270.76\ N$

Now, we know the work done is force times displacement.

So,

$W=P\times d\\W=270.76\times 10.4=2815.90\ J\\W=2.81\ kJ$

So, the total work done by the mover to displace 46 kg crate by 10.4 meters 2.81 kJ.