Home
Class 11
PHYSICS
Two masses of 5kg and 3kg are suspended ...

Two masses of `5kg` and `3kg` are suspended with help of massless inextensible strings as shown in figure. Calculate `T_(1)`and `T_(2)` when whole system is going upwards with acceleration `=2m//s^(2) (use g = 9.8 ms^(-2))`.

Text Solution

Verified by Experts

Given `m_(1) = 5` kg , `m_(2) = 3` kg
g = `9.8 m//s^(2)` and a = 2 m/ `s^(2)`

For the upper block
`T_(1) - T_(2) - 5 g = 5a `
`implies " " T_(1) - T_(2) = 5(g +a) " " .....(i) `
For the lower block `" " T_(2) - 3g = 3a`
`implies " " T_(2) = 3(g+a) = 3(9.8 + 2) = 35.4 N`
From Eq. (i) ` " " T_(1) = T_(2) + 5(g +a)`
= `35.4 + 5(9.8 + 2) = 94.4 N`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • LAWS OF MOTION

    NCERT EXEMPLAR|Exercise Long answer Type Questions|9 Videos

  • LAWS OF MOTION

    NCERT EXEMPLAR|Exercise Very short answer type Questions|12 Videos

  • KINETIC THEORY

    NCERT EXEMPLAR|Exercise Multiple Choice Questions (MCQs)|31 Videos

  • MECHANICAL PROPERTIES OF FLUIDS

    NCERT EXEMPLAR|Exercise Long Answer Type Questions|3 Videos

Similar Questions

Explore conceptually related problems

Two masses of 5 kg and 3 kg are suppended with help of massless inextensible strings as shown in Calculate T_(1) and T_(2) when whole system is going upwards with acceleration =2ms^(2) (us g= 9.8ms^(-2) ) .

Two masses of 5 kg and 3 kg are suspended with help of mass less inextensible string as shown. Calculate T_1 and T_2 when system is going upwards with acceleration m//s^2. ("Use g"= 9.8m//s^2) .

Knowledge Check

  • A block of mass 10 kg is suspended by three strings as shown in figure . Tension T_(2) is

    A
    100N
    B
    `(100)/(sqrt(3))N`
    C
    `sqrt(3)xx100N`
    D
    `50sqrt(3)N`

  • Two bodies of mass 3kg and 4kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is ( g=9.8m//s^(2) )

    A
    `4.9 m//s^(2)`
    B
    `2.45 m//s^(2)`
    C
    `1.4 m//s^(2)`
    D
    `9.5 m//s^(2)`

  • Two bodies of mass 5 kg and 7 kg are suspended at the ends of massless string passing over frictionless pulley. The acceleration of the system.

    A
    `1.6 m//s^(2)`
    B
    `10 m//s^(2)`
    C
    `9.8 m//s^(2)`
    D
    `7 m//s^(2)`

    • Similar Questions

    Explore conceptually related problems

    The masses M_(1), M_(2) and M_(3) are 5, 2 and 3 kg respectively. These have been joined using massless, inextensible pieces of strings as shown in the figure. If the whole system is going upward with an acceleration of 2ms^(-2) , then the value of tensions T_(1), T_(2) and T_(3) respectively are ["take "g=9.8ms^(-2)]

    The masses m_(1) m_(2) and m_(3) of the three bodies shown in fig . Are 5 , 2 and 3 kg respectively Calculate the valuse of tension T_(1) T_(2) and T_(3) when (i) the whole system is going upward with an acceleration of 2 m//s^(2) (ii) the whole system is stationary (g=9.8 m//s^(2)) . .

    A block of mass 10 kg is suspended with two strings as shown in the fig. Find the tensions T_(1) and T_(2) in the two string.

    Two blocks of masses 2 kg and 4 kg are hanging with the help of massless string passing over an ideal pulley inside an elevator. The elevator is moving upward with an acceleration (g)/(2) . The tension in the string connected between the blocks will be (Take g=10m//s^(2) ).

    Two bodies of mass 3 kg and 4 kg are suspended at the ends of massless string passing over a frictionless pulley. The acceleration of the system is (g=9.8m//s^(2))

    Home
    Profile