The crate is in equilibrium. Newton's second law gives
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0
where
• <em>n</em> = magnitude of the normal force
• <em>mg</em> = weight of the crate
• <em>p</em> = mag. of push exerted by movers
• <em>f</em> = mag. of kinetic friciton, with <em>f</em> = 0.60<em>n</em>
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It follows that
<em>p</em> = <em>f</em> = 0.60<em>mg</em> = 0.60 (43.0 kg) <em>g</em> = 252.84 N
so that the movers perform
<em>W</em> = <em>p</em> (10.4 m) ≈ 2600 J
of work on the crate. (The <em>total</em> work done on the crate, on the other hand, is zero because the net force on the crate is zero.)