during a storm a crate is sliding a cross horizontal road . which is oily and its moves a displacement 3m. why a steady wind pushe against a creat f = (2i-0.6j)n find how much work does the wind on the crate to finally stops within 3m
Answers
Answer:
(a) -6.0 J;
(b) 4.0 J.
Explanation:
(A) Because we can treat the crate as a particle and because the wind force is constant (“steady”)
in both magnitude and direction during the displacement to calculate the work we can use equation
= ⃗ ∙ ⃗
= [(2.0N)⃗i + (−6.0N)⃗j] ∙ [(−3.0m)⃗i ] =
= (2.0N)(−3.0m)⃗i⃗i + (−6.0 )(−3.0 )⃗i⃗j =
=(-6.0 J)(1)+0=-6.0 J
Thus, the force does a negative 6.0 J of work on the crate, transferring 6.0 J of energy from the
kinetic energy of the crate.
(b) Because the force does negative work on the crate, it reduces the crate’s kinetic energy.
Using the work–kinetic energy theorem we have
= + = 10 J + (−6.0 J) = 4.0 J.
Less kinetic energy means that the crate has been slowed.
Answer: (a) -6.0 J;
(b) 4.0 J.
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