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A sled is dragged along a horizontal path at a constant speed of 1.50 m/s by a rope that is inclined at an angle of 30.0° with respect to the horizontal. The total weight of the sled is 470 N. The tension in the rope is 230 N. How much work is done by the rope on the sled in a time interval of 10.0 s?

Respuesta :

Answer:2987.7 J

Explanation:

Given

sled weight is [tex]W=470 N[/tex]

Tension in rope [tex]T=230 N[/tex]

Time interval [tex]t=10 s[/tex]

velocity of sled [tex]v=1.5 m/s[/tex]

here speed is equal to velocity which is constant therefore acceleration is zero

thus [tex]F_{net}=0[/tex]

Distance traveled in [tex]t=10 s[/tex]

[tex]x=v\times t=1.5\times 10=15 m[/tex]

considering surface to be friction less

thus [tex]T\cos 30[/tex] will do work on sled

[tex]W=F\cdot x[/tex]

[tex]W=T\cos 30\cdot x[/tex]

[tex]W=230\cos 30\times 15[/tex]

[tex]W=2987.78 J\approx 2.987 kJ[/tex]

onyebuchinnaji

The work-done by the rope on the sled in a time interval is 2,987.7 J.

The given parameters;

  • constant speed of the sled, u = 1.5 m/s
  • angle of inclination of the sled, Ф = 30°
  • the total weight of the sled, W = 470 N
  • tension on the rope, T = 230 N
  • time of motion, t = 10.0 s

The distance traveled by the sled at the given time period is calculated as follows;

x = ut

x = 1.5 x 10

x = 15 m

The work-done by the rope on the sled in a time interval is calculated as;

Work-done = Tcos(Ф)x

Work-done = 230 x cos(30) x 15

Work-done = 2,987.7 J.

Thus, the work-done by the rope on the sled in a time interval is 2,987.7 J.

Learn more here:https://brainly.com/question/19498865

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