Question

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

Answer 1

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

Answer 2

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