Let $μ$ be the coefficient of friction between blocks of mass $m$ and $M$ . The pulleys are frictionless and strings are massless. Acceleration of mass $m$ is:

\frac{\sqrt{5} m g}{M + \sqrt{5} m + 2 μ m}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 m g}{m + 5 M + 2 μ m}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 \sqrt{5} m g}{M + 5 m + 2 μ m}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{5 \sqrt{2} m g}{m + \sqrt{5} m + \sqrt{2} μ m}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{2 \sqrt{5} m g}{M + 5 m + 2 μ m}$
From FBD,
for mass $m$ : $m a_{y} = m g - T - f = m g - T - μ N . . . . . (1)$
and $N = m a_{x}$
for mass $M$ : $M a_{x} = 2 T - N = 2 T - m a_{x} . . . . . (2)$
solving (1) and (2) , we will get $a_{x} = \frac{2 m g}{M + 5 m + 2 μ m}$
and $a_{y} = \frac{4 m g}{M + 5 m + 2 μ m}$
The acceleration of mass $= a = \sqrt{a_{x}^{2} + a_{y}^{2}} = \frac{2 \sqrt{5} m g}{M + 5 m + 2 μ m}$

Suggest Corrections

Similar questions

Q. Two blocks of mass m and 2m are connected as shown in figure. A force F is acting on the block of mass 4m. Pulley is massless and string is light and inextensible. All the surfaces are smooth.
Contact force between ground and the 4m mass under the previous problem condition is

Q. A cart of mass

M

moves with acceleration

a

as shown in the figure. The minimum force required to hold a block of mass

m

together is

(

friction coefficient between two blocks is

μ)

Q. If

m - \frac{1}{m} = 5

, then find the value of

m^{2} + \frac{1}{m^{2}}

A plate of mass M is placed on a horizontal frictionless surface (see figure) and a body of mass m is placed on this plate. The coefficient of dynamics friction between this body and the plate is μ. If a force 2 μmg is applied to the body of mass m along the horizontal, the acceleration of the plate will be

In the pulley arrangement shown below, the pulley $P_{2}$ is movable. Assuming the coefficient of friction between m and surface to be $μ$ , the maximum value of M for which m is at rest is

Join BYJU'S Learning Program

Let μ be the coefficient of friction between blocks of mass m and M. The pulleys are frictionless and strings are massless. Acceleration of mass m is:

The correct option is C 2√5mgM+5m+2μmFrom FBD, for mass m : may=mg−T−f=mg−T−μN.....(1)and N=maxfor mass M: Max=2T−N=2T−max.....(2)solving (1) and (2) , we will get ax=2mgM+5m+2μmand ay=4mgM+5m+2μmThe acceleration of mass =a=√a2x+a2y=2√5mgM+5m+2μm

Let $μ$ be the coefficient of friction between blocks of mass $m$ and $M$ . The pulleys are frictionless and strings are massless. Acceleration of mass $m$ is: