A 4.50 kg block initially at rest is pulled to the right along a horizontal, fri

Question

A 4.50 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a rope as shown in the figure. The rope has a tension of 15.0N, and is held at an angle of 15.0degree below the horizontal. The block is pulled 5.0 m along the floor. Find the work due to the rope, gravity, and the normal force on the block. Using the Work-Kinetic Energy theorem to find the speed after the 5.0 m. It takes 1.76s to cover this distance. What is the power provided by the rope? W = Fdcostheta

Explanation / Answer

here,
m = 4.50 kg
Tx = 15 Cos15 = 15*0.96 = 14.4 N
Ty = 15Sin15 = 15*0.258 = 3.88 N

Force due to gravity :
Fg = m*g = 4.50 * 9.8
Fg = 44.1 N

Normal Force =

N = mg
N = 44.1 N

Force on block due to pulling

F = Tx = 14.4 N

Work done = Force * Displacement *CosA

a)

Work done to rope :
W = Fx.d.CosA
as force and displacement are parallel to each other therefore

A= o degrees or Cos0 = 1

Wr = 14.4 * 5
Wr = 72 J

Work done due to normal and gravity will zero as there is no displacement in y direction.

B)

From work Energy theoram we have :
Word done in pulling = kinetic energy gained by block
72 = 0.5 *m * v^2
v = sqrt(72*2/4.5)
v = 5.657 m/s

C)
power = work done/time
P = 72/1.76
P = 40.909 Watt

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