A hoop of radius 2 m weighs 100 kg. It rolls along a horizontal floor so that its centre of mass has a speed of 20 cm/s. How much work has to be done to stop it?
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Hey mate,
◆ Answer - 4 J.
◆ Explaination-
# Given-
r = 2 m
m = 100 kg
v = 20 cm/s = 0.2 m/s
# Solution-
Translational K.E. of hoop-
KE(T) = 1/2 mv^2
Rotational K.E. of hoop-
KE(R) = 1/2 Iω^2
For hoop, I = mr^2
KE(R) = 1/2 mr^2ω^2
KE(R) = 1/2 mv^2
Work done for stopping the hoop is equal to total energy of the hoop.
W = Et = KE(T) + KE(R)
W = 1/2 mv^2 + 1/2 mv^2
W = mv^2
W = 100×(0.2)^2
W = 100×0.04
W = 4 J
Work done in stopping the hoop is 4 J.
Hope that was useful..
◆ Answer - 4 J.
◆ Explaination-
# Given-
r = 2 m
m = 100 kg
v = 20 cm/s = 0.2 m/s
# Solution-
Translational K.E. of hoop-
KE(T) = 1/2 mv^2
Rotational K.E. of hoop-
KE(R) = 1/2 Iω^2
For hoop, I = mr^2
KE(R) = 1/2 mr^2ω^2
KE(R) = 1/2 mv^2
Work done for stopping the hoop is equal to total energy of the hoop.
W = Et = KE(T) + KE(R)
W = 1/2 mv^2 + 1/2 mv^2
W = mv^2
W = 100×(0.2)^2
W = 100×0.04
W = 4 J
Work done in stopping the hoop is 4 J.
Hope that was useful..
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