# 1)a lift is moving down with an acceleration equal to the acc due 2 gravity.abody of massM kept on the floor of the lift is pulled horizontally.if coeff of fric is m den the frictional resistance is 2)2 blocks of mass 4kg and 2 kg are connected by a heavy spring and placed on rough horizontal surface .the 2 kg block is pulled with a constant force F.the coeff of fric bet the blocks and the ground is 0.5.wat is the value of F so that tension in the string is constant throughout during the motion of the blocks

1
by aimanakhtar799
i want the solution

2014-10-11T21:07:07+05:30

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1)
mass = M
Acceleration of life moving down = g    Coeff of friction is μ.
The mass M and the lift are both travelling with the same speed and acceleration. So there is no weight force acting on the floor of the lift. There is no normal reaction from the lift. they are just one over another, that is all.

If the mass is pulled along the horizontal direction,  Frictional resistance force = 0.
Because the contact forces are zero.  If the body is pushed or pulled in another direction, then there will be friction.

2) you mention spring and  string ???      I assume it as a string.

masses 4 kg and 2 kg                 2 kg is pulled by Force F
coeff of friction = μ = 0.5

Let the tension in the string be T. If T is constant, then the string is tight and the two blocks at the two ends are moving at the same speed and acceleration. Let them move with an acceleration a.  Actually a = 0, for tension to be same through out the string.

equation for 2 kg :        m a = F - T - μ m g     =>  a = (F/2 - T/2 - g/2)
So the equation for 4 kg :   M a = T - μ M g   =>  a = T/4 - g/2

so  F/2 - T/2 - g/2 = T/4 - g/2
3T/4 = F/2 - g/2 = F / 2
T = 2 F / 3

so     a = F/6 - g/2              F = (6 a + 3 g )    =  (6 a + 30) Newtons

If both blocks are moving with a uniform velocity, a = 0
F = 30 Newtons, if blocks move with uniform velocity and string has a constant tension throughout.