Laws of MotionHard

Question

In the arrangement show in figure m1=1kg, m2=2kg. pulleys are mass less and strings are light. For What value of M the mass m1moves with constant velocity (Neglect friction)

Options

A.6 kg
B.4 kg
C.8 kg
D.10 kg

Solution

T = M ×.........(i)
20 - 2 × a
20 - = 2a
& = 1 × g ⇒ T = 20 N
20 - 10 = 2a ⇒ a = 5 m/s2
20 - = 2 × 5 ⇒  m =
M = 8 kg

Create a free account to view solution

View Solution Free
Topic: Laws of Motion·Practice all Laws of Motion questions

More Laws of Motion Questions