Question

2. A block with mass M1 rests on a frictionless table. It is connected by a massless string to a block with mass M2, which hangs below the edge of the table. The system is released from rest at time t = 0. Find the distance block M1 moves in time t. You may assume that the string passes over a massless, frictionless pulley at the edge of the table to assist your calculations.

Answer 1

Forces on M₂ include:

1.force due to gravity,mg,where m is mass and g is gravitational acceleration

here m=M₂ , so force due to gravity=M₂*g (downwards)

2.Force due to tension in the string= T (upwards)

Let the acceleration of this object be a (downwards)

So, using newton's second law,F=ma, where F is the net force acting on an object, m is its mass, a is its acceleration, we get,

M₂*g-T=M₂*a=>T=M₂(g-a).........................equation 1

Horizontal forces on M₁ include:

1.Force due to string ,T

magnitude of acceleration of M₁= magnitude of acceleration of M₂(since,they are connected by an inextensible string whose length is constant)

so, magnitude of acceleration of M₁=a

So, using newton's second law, T=M₁*a

Substituting the value of T from equation 1, we get

M₂(g-a)=M₁*a=>M_2*g-M_2*a=M₁*a=>(M₁+M₂)a=M₂*g=>a=M₂*g/(M₁+M₂)

Now, under uniform acceleration, s=ut+(1/2)*at² ,where s is displacement, u is initial velocity, t is time interval,a is acceleration.

In given problem,the system starts from rest, so u=0

Hence, distance traveled s=(1/2)*a*t²=(1/2)*[M₂*g/(M₁+M₂)]*t²=M₂*gt²/[2(M₁+M₂)]