Question

The block has mass M and rests on a surface for which the coefficient of static and kineticfriction are mu_(s) and mu_(k) respectively.A force F=kt^(2) is applied to the cable.Velocity of block at t=2sec is vm/s .Then value of v is.Given :M=24kg k=60N/s^(2) mu_(s)=0.5 mu_(k)=0.4​

Answer 1

Explanation:

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Answer 2

Given:

Mass of block $=24kg$

Force equation $F=kt^{2}$

The constant $k=60N/s^{2}$

Coefficient of static friction μ$_{s} =0.5$

Coefficient of kinetic friction μ$_{k}=0.4$

To find:

The velocity of the block at $t=2$ $sec$

Solution:

Step 1

We have been given that initially a block of mass $m=24kg$ is resting on a horizontal surface until when a force of $F=kt^{2}$ is applied on the block.

Initially when the block is not moving, it exerts a force on the ground  which is technically equal to its own weight .Hence normal reaction to this force is,

$R=mg$

Now,

If we see the $FBD$ (Free body diagram) of the block when force is starting to build up, we see the forces acting on the block to make it move are

$F_{ext}-f_{s} \geq 0$

$F_{ext}\geq f_{s}$

The external force applied on the block to make it move should be greater than or equal to the static friction.

Step 2

Now,

When the block begins to move under the effect of the external force, at this time, the major forces acting on the block will be

$F_{ext}-f_{k}=ma$

Substituting the values, we get

$kt^{2} -$ μ$_{k}R=ma$

$60(2)^{2} -0.4(24)(10)=24a$

$240-96=24a$

$24a=144$

$a=6m/s^{2}$

Hence, the block moves with an acceleration of 6 m/s².

Step 2

Now,

We know, the block starts from rest, the velocity at 2 s will be given by,

$v=u+at$

We have,

$u=0m/s$    $;$ $a=6m/s^{2}$    $;$ $t=2s$

Substituting the given values, we get

$v=0+6(2)$

$v=12m/s$

Final answer:

Hence, the velocity of block at t = 2 s is 12 m/s.