NEET2015PhysicsLaws of MotionActual
A block A of mass m 1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m 2 is suspended. The coefficient of kinetic friction between the block and the table is μ k . When the block B is sliding on the table, the tension in string is:
Options
- Am 2 + μ k m 1 g m 1 + m 2
- Bm 2 - μ k m 1 g m 1 + m 2
- Cm 1 m 2 1 + μ k g m 1 + m 2
- Dm 1 m 2 1 - μ k g m 1 + m 2
Correct answer
C. m 1 m 2 1 + μ k g m 1 + m 2
Step-by-step solution
From the figure, ( aligned & m₂ g-T=m₂ a (i) & T- _k m₁ g=m₁ a (ii) & multiply eqn (i) with m ₁ and eqn (ii) with m ₂ & m₁ m₂ g-T m₁=m₁ m₂ a (iii) & T m₂- _k m₁ m₂ g=m₁ m₂ a (iv) & From eqn (iii) and (iv), we get, & m₁ m₂ g-T m₁=T m₂- _k m₁ m₂ g & m₁ m₂ g (1+ _k )=T (m₁+m₂ ) & T= m₁ m₂ g (1+ _k ) m₁+m₂ aligned )