Question

The block has mass M and rests on a surface for
which the coefficients of static and kinetic friction
are ps and uk respectively. A force F = kt2 is applied
to the cable. Velocity of block at t = 2 sec is v m/s.
Then value of v is. Given : M = 24 kg, k = 60 N/s2,
Hs = 0.5, k = 0.4
M
F​

Answer 1

Answer:

Physics Q 9 Answer is 19.6.

Answer 2

Given:

Mass $=24kg$

$F=kt^{2}$

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

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:

We have been given that a block of mass $m=24kg$ initially rests on a horizontal surface when an external force of $kt^{2}$ is applied on it which gives it a certain velocity.

For the body to leave its state of rest and start moving, an external force, greater than its static friction has to be supplied to it to make it moving.

When the block once starts moving, the free body diagram $FBD$ of the block shows the following forces acting on it.

1. The normal reaction of the block of the mass $(R=mg)$.
2. The external force $(F_{ext} =kt^{2})$.
3. The frictional force on the block. ( $f_{k} =$ μ$_{k}R$ )    

Now, if the external force is greater than the frictional force on the block, the block begins to move with an acceleration $a$.

Hence,

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

$kt^{2} -$ μ$_{k}R=ma$

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

$240-96=24a$

$a=\frac{144}{24}=6m/s^{2}$

Now, we have

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

$v=u+at$

$v=0+6(2)$

$v=12m/s$

Final answer:

Hence, the block moves with a velocity of 12 m/s.