A horizontal force of 840 N is needed to drag a crate across a horizontal floor with a constant speed. You drag the crate using a rope held at an angle of 28°.
What force do you exert on the rope?
How much work do you do on the crate when moving it 22 m?
If you complete the job in 8 s, what power is developed (in kW)?
How would you solve these problems? With what formulas?
2006-10-31 07:16:13 · 4 answers · asked by Isabel G 1 in Science & Mathematics ➔ Physics
If F is the force you exert on the rope, then:
F*cos(28) = 840, F = 840/cos(28), F = 840/0.883 = 951.4 N (approx)
work = force*distance = 840*22 = 18480 J
power = work/time = 18480/8 = 2310 W or 2.31 kW
2006-10-31 07:23:53 · answer #1 · answered by Dimos F 4 · 0⤊ 0⤋
You need to break up the force on the rope into horizontal and vertical components, set the horizontal component to 840N and calculate the force on the rope as
Fr = 840/cos(28) = 951.36N
Work is force thru a distance so the work done is
840N*22m which is 18,480J.
Power is work/second so
18,480J/8s = 2,310W or 2.31KW
Doug
2006-10-31 15:26:30 · answer #2 · answered by doug_donaghue 7 · 0⤊ 0⤋
The force on the rope must be 840N/cos(28°)=951,36N
The work is the force * 22m = 951.36*22=20929,92 Joules
The power is work/time= 2616,24 W
2006-10-31 15:27:52 · answer #3 · answered by Claude J 2 · 0⤊ 0⤋
hi
i think the answer is
force=M.a
since its inclined 28degrees tension on he rope is 840.cos28degrees
and work=F.S i.e force into displacement okay so its 840.cos28.22m okay
power =work/time so its 840.cos28.22/8 seconds
by the way "." = product
2006-10-31 15:27:11 · answer #4 · answered by sunny 1 · 0⤊ 0⤋