Question

A block of mass 5 kg is lying on a rough horizontal surface. The coefficients of static and
kinetic friction between the block and the surface respectively are 0.7 and 0.5. A horizontal
force just sufficient to move the block is applied to it. If the force continues to act even after the
block has started moving, the acceleration of block in m/s2 will be take g = 10 ms?).

Answer 1

Force applied, F= μsmg

=0.7x 5x 10= 35N

Frictional force when block starts moving,

f = μk mg = 0.5x5x10= 25N

ma = Fapplied - f

= 35 - 25 = 10N

a = 10N/ 5kg= 2m/s²

Answer 2

Given:

A block of mass 5 kg is lying on a rough horizontal surface. The coefficients of static and

kinetic friction between the block and the surface respectively are 0.7 and 0.5. A horizontal

force just sufficient to move the block is applied to it.

To find:

the acceleration of block in m/s².

Calculation:

The critical force required to move the block against static friction is :

$\therefore \: \rm{F = limiting \: friction}$

$= > \: \rm{F = \mu_{s} \times N }$

$= > \: \rm{F = \mu_{s} \times (mg) }$

$= > \: \rm{F = 0.7 \times (5 \times 10) }$

$= > \: \rm{F = 35 \: N }$

Now , kinetic friction be f ;

$\therefore \: \: \rm{f = \mu_{k} \times N }$

$= > \: \: \rm{f = 0.5 \times (5 \times 10) }$

$= > \: \: \rm{f = 25 \:N }$

Let acceleration be a ;

$= > \: \rm{F - f= ma }$

$= > \: \rm{35 - 25= 5a }$

$= > \: \rm{5a = 10 }$

$= > \: \rm{a = 2 \: m {s}^{ - 2} }$

So , final answer is:

$\boxed{ \bf{acceleration = 2 \: m {s}^{ - 2} }}$