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A 20.0-kg cate sits at rest at the bottom of a 15.0 m-long
Chapter 3, Problem 65P(choose chapter or problem)
A 20.0-kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at \(34.0^{\circ}\) above the horizontal. A constant horizontal force of 290 N is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 N.
(a) What is the total work done on the crate during its motion from the bottom to the top of the ramp?
(b) How much time does it take the crate to travel to the top of the ramp?
Questions & Answers
QUESTION:
A 20.0-kg crate sits at rest at the bottom of a 15.0-m-long ramp that is inclined at \(34.0^{\circ}\) above the horizontal. A constant horizontal force of 290 N is applied to the crate to push it up the ramp. While the crate is moving, the ramp exerts a constant frictional force on it that has magnitude 65.0 N.
(a) What is the total work done on the crate during its motion from the bottom to the top of the ramp?
(b) How much time does it take the crate to travel to the top of the ramp?
ANSWER:Step 1 of 3:
a)The normal force is perpendicular to displacement does no work so The normal force does no work.
The other work is
\(W=(F \cos ) s F=(290 N)(\cos 34) 15 \mathrm{~m}=3606 \mathrm{~J} \mathrm{~W} \mathrm{mg}=(F \sin ) \mathrm{s} 0=(\mathrm{mg})(\sin 34) 15 \mathrm{~m}=20 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s} \times 0.559 \times 15 \mathrm{~m}=1664 \mathrm{JW}=\mathrm{f}(15 \mathrm{~m}) \mathrm{f}=65 \mathrm{~N} \times 15 \mathrm{~m}=975 \mathrm{~J}\)
The total work is W total W +FW mg + W f = 3606 J 1664 J 975 J = 987 J