What is the acceleration of the block and trolley system shown in the figure $5.12 (a)$ , if the coefficient of kinetic friction between the trolley and the surface is $0.04 ?$ . What is the tension in the string. (Take $g = 10 m s^{- 2})$ Neglect the mass of the string.

Open in App

Solution

Acceleration of both trolley and block $a$ will be same. As we can see $(f_{s})_{m a x} = 0.04 \times 20 \times 10$
$= 8 N < 30 N$
$∴$ Block will go down.
$30 - T = 3 a ⟶ (i)$
$T - f_{k} = 20 a ⟶ (i i)$
on solving $(i)$ and $(i i)$ we get
$30 - f_{k} = 23 a$
$30 - 8 = 23 a$
$a = \frac{22}{23} m / s^{2}$
$a = 0.956 m / s^{2}$
$T = 30 - 3 a$
$= 30 - 3 \times \frac{22}{23}$
$T = 27.13 N$

Suggest Corrections

Similar questions

Q. If coefficient of kinetic friction is

μ_{k} = 0.04

between the surface and trolley then the acceleration of trolley-block system as shown in diagram will be

(g = 10 m / s^{2})

Q. A trolley of mass 20 kg is attached to a block of mass 4 kg by a massless string passing over a frictionless pulley as shown in the figure. If the coefficient of kinetic friction between trolley and the surface is 0.02, then the acceleration of the trolley and block system is:
(Take g = 10

m s^{- 2}

)

A trolley of mass M is attached to a block of mass m by a string passing over a frictionless pulley as shown in the figure. If the coefficient of friction between the trolley and the surface below is $μ,$ what is the acceleration of the trolley and the block system when they are released?