A pendulum string with the length of 105 cm has the 325 g bob attached to it. At the lowest point of the swing the pendulum bob moves with the speed of 2.15 m/s. Determine: a) the centripetal acceleration of the pendulum bob at this point; b) the force of tension in the string at this point; c) the kinetic energy of the pendulum bob at this point;

(a) 4.4 m/s²

(b) 4.615 N.

(c) 0.751 J

Explanation:

(a)

Using,

a' = v²/r Equation 1

Where a' = centripetal acceleration, v = speed of the pendulum bob, r = radius or length of the pendulum bob.

Given: v = 2.15 m/s, r = 105 cm = 1.05 m

Substitute into equation 1

a' = 2.15²/1.05

a' = 4.4 m/s².

(b) The force of tension in the string = Tangential force + weight of the bob.

T = ma'+mg Equation 2

Where T = Force of tension in the string, m = mass of the bob, g = acceleration due to gravity.

Given: m = 325 g = 0.325 kg, a' = 4.4 m/s², g = 9.8 m/s²

Substitute into equation 2

T = 0.325×4.4+0.3325×9.8

T = 1.43+3.185

T = 4.615 N.

(c) Kinetic energy of he bob at that point = 1/2mv²

Ek = 1/2mv² Equation 3

Where Ek = kinetic energy of the bob

Given: m = 0.325 kg, v = 2.15 m/s

Substitute into equation 3

Ek = 1/2(0.325)(2.15²)

Ek = 0.751 J

c) power lifting, will be your answer.

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