A block of mass 40 kg slides over a surface, when a mass of 4 kg is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (Given g=10 ms -2.)

Question

A block of mass 40 kg slides over a surface, when a mass of 4 kg is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (Given g=10 ms -2.) (A) 1 ms -2 (B) 1 / 5 ms -2 (C) 4 / 5 ms -2 (D) 8 / 11 ms -2

Tardigrade Admin · Accepted Answer

For 4 kg block 4 g - T =4 a For 40 kg block T -40 g × 0.02=40 a Adding both eq. 40-8=44 a a=(32/44)=(8/11) m / s 2

$1 m s^{- 2}$

$1/5 m s^{- 2}$

$4/5 m s^{- 2}$

$8/11 m s^{- 2}$

For $4 k g$ block
$4 g - T = 4 a$
For $40 k g$ block
$T - 40 g \times 0.02 = 40 a$
Adding both eq.
$40 - 8 = 44 a$
$a = \frac{32}{44} = \frac{8}{11} m / s^{2}$

Q. A block of mass 40kg slides over a surface, when a mass of 4kg is suspended through an inextensible massless string passing over frictionless pulley as shown below. The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (Given g=10ms−2.)

1ms−2

1/5ms−2

4/5ms−2

8/11ms−2

For 4kg block 4g−T=4a For 40kg block T−40g×0.02=40a Adding both eq. 40−8=44a a=4432​=118​m/s2

$1 m s^{- 2}$

$1/5 m s^{- 2}$

$4/5 m s^{- 2}$

$8/11 m s^{- 2}$

For $4 k g$ block
$4 g - T = 4 a$
For $40 k g$ block
$T - 40 g \times 0.02 = 40 a$
Adding both eq.
$40 - 8 = 44 a$
$a = \frac{32}{44} = \frac{8}{11} m / s^{2}$