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Standard XII

Physics

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A block of ma...

Question

A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and g = 9.8 $m / s^{2}$ , then the acceleration of the block is

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Solution

Given : $m = 5 k g$ $F = 24 N$ $μ_{k} = 0.4$
Force $m a = F - f_{k}$
Or $m a = F - μ_{k} m g$
Or $5 a = 24 - 0.4 \times 5 \times 9.8$
$⟹ a = 0.88 m / s^{2}$

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Similar questions

Q. A block of mass

5 kg

is on a rough horizontal surface at rest. Now a force(

F

) of

24 N

is imparted to it with negligible impulse. If the coefficient of kinetic friction is

0.4

and

g = 9.8 {m/s}^{2}

, then the acceleration of the block is

Q. A block of mass

5 k g

is on a rough horizontal surface and is at rest. Now a force of

24 N

is imparted to it with negligible impulse. If the coefficient of kinetic friction is

0.4

and

g = 9.8 m / s^{2}

, then the acceleration of the block is

A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and g=9.8 $m / s^{2}$ , then the acceleration of the block is

Q. A block of mass

5 k g

is on a rough horizontal surface and is at rest. Now a force of

24 N

is imparted to it with negligible impluse. If the coefficient of kinetic friction is

0.4

and

g = 9.8 m / s^{2}

then the acceleration of the block is

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A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and g = 9.8 m/s2, then the acceleration of the block is

Given : m=5 kg F=24 N μk=0.4Force ma=F−fkOr ma=F−μkmgOr 5a=24−0.4×5×9.8⟹ a=0.88 m/s2

A block of mass 5 kg is on a rough horizontal surface and is at rest. Now a force of 24 N is imparted to it with negligible impulse. If the coefficient of kinetic friction is 0.4 and g = 9.8 $m / s^{2}$ , then the acceleration of the block is

Given : $m = 5 k g$ $F = 24 N$ $μ_{k} = 0.4$
Force $m a = F - f_{k}$
Or $m a = F - μ_{k} m g$
Or $5 a = 24 - 0.4 \times 5 \times 9.8$
$⟹ a = 0.88 m / s^{2}$