Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure.
The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is
23.3 kg
43.3 kg
10.3 kg
18.3 kg
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Answered by
6
Answer:
✏ The correct option is A.
Explanation:
m1 = 5 kg and m2 = 10 kg
As the system is at rest, T = 50 N
50 – T = 5 × a
T – 0.15(m + 10) g = (10 + m)a
a = 0 for rest
50 = 0.15(m + 10) × 10
5 = 3/20(m + 10)
100/3 = m + 10
✒ m = 23.3 kg
Answered by
4
Explanation:
m1 = 5 kg and m2 = 10 kg
As the system is at rest, T = 50 N
50 – T = 5 × a
T – 0.15(m + 10) g = (10 + m)a
a = 0 for rest
50 = 0.15(m + 10) × 10
5 = 3/20(m + 10)
100/3 = m + 10
✒ m = 23.3 kg
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