To push a 52-kg crate across a floor, a worker applies a force of 190 N, directed 22° below the horizontal. As the crate moves 3.3 m, how much work is done on the crate by (a) the worker, (b) the force of gravity, and (c) the normal force of the floor on the crate?
Answers
Explanation:
(a) the work is done on the crate by the worker , substituent 190N for F and 3.3m for S and 22⁰ for ø in the equation W = FS cos ø,
W = FS cos ø
= ( 190 N ) ( 3.3 m ) cos ( 22⁰ ).
= ( 580 N.m ) ( 1 J ÷ 1 N.m ).
= 580 J.
(b) the force gravity is perpendicular to the displacement of the crate . so, ø = 90⁰ for this case .
Thus , W = Fs cos ø
= Fs cos 90⁰
= 0 ( Since , cos 90⁰ = 0 )
(c) the normal force of the force on the crate is perpendicular to the displacement of crate . so again ø = 90⁰ for this case.
Thus , W = Fs cos ø
= Fs cos 90⁰
= 0 ( since cos 90⁰ = 0 )
therefore, the work W is done on th crate by the normal force on the floor on the crate would be zero.
Answer:
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