A small rock with mass 0.15 kg is fastened to a massless string with length 0.80

Question

A small rock with mass 0.15 kg is fastened to a massless string with length 0.80 m to form a
pendulum. The pendulum is swinging so as to make a maximum angle of 30° with the vertical. Air
resistance is negligible.
a) What is gravitational potential energy of the pendulum when it makes an angle of 30° with the
vertical? (2 points)
b) What is the speed of the rock when the string passes through the vertical position? (3 points)
c) What is the tension in the string as it passes through the vertical position? (3 points)
d) What is the work done by the tension in the string when the pendulum swings from the vertical
position to making an angle of 30° with the vertical? (2 points)

Explanation / Answer

a)Height of pendulum above bottommost position = 0.8*(1-cos30)=0.107m

Potential energy = mgh = 0.15*9.8*0.107=0.157 Joule

b)The speed of rock when it passes through vertical postion = sqrt(2*9.8*0.107)=1.45m/s

c)tension in the string as the rock passes vertical position = mg+mv2/l = (0.15*9.8)+(0.15*1.452/0.8)=1.86N

d)Since there is conservation of energy, the work done by tension in string =0

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