1.A 43.0 -kg child swings in a swing supported by two chains, each 3.04 m long.

Question

1.A 43.0-kg child swings in a swing supported by two chains, each 3.04 m long. The tension in each chain at the lowest point is 356 N.

A 43.0-kg child swings in a swing supported by two chains, each 3.04 m long. The tension in each chain at the lowest point is 356 N. Find the child's speed at the lowest point. Find the force exerted by the seat on the child at the lowest point. (Ignore the mass of the seat.) A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in the figure below. The length of the arc ABC is 270 m, and the car completes the turn in 31.0 s. What is the acceleration when the car is at B located at an angle of 35.0 degree ? Express your answer in terms of the unit vectors and m/s2 m/s2 . Determine the car's average speed. m/s Determine its average acceleration during the 31.0-s interval. m/s2 m/s2

Explanation / Answer

1

a)

2T = mg + mv^2 /r

2*356 = 43*9.81 + 43*v^2 / 3.04

v = 4.529 m/s

b)

N = 2T = 2*356 = 712 N

2)

s = r*theta

270 = r*(pi/2)

r = 171.974 m

v = s/t = 270 / 31 = 8.7097 m/s

a = v^2 /r

a = 8.7097^2 / 171.974

a = 0.4411 m/s^2

a = 0.4411*cos35 i - 0.4411*sin35 j

a = 0.3613 i + (-0.253) j

Hence,

a_x = 0.3613 m/s^2

a_y = -0.253 m/s^2

b)

v = 8.7097 m/s

c)

In 31 s, Vx reduces from 8.7097 m/s to 0. And Vy increases from 0 to 8.7097 m/s.

Acceleration, a_x = (0 - 8.7097) / 31 = -0.281 m/s^2

a_y = (8.7097 - 0) / 31 = 0.281 m/s^2

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