A body moves on a rough horizontal surface with acceleration $a$ when it is subjected to a horizontal force. If the applied force is doubled, it moves with acceleration
equal to $a$
equal to $2 a$
less than $2 a$
greater than $2 a$

Question

A body moves on a rough horizontal surface with acceleration $a$ when it is subjected to a horizontal force. If the applied force is doubled, it moves with acceleration
equal to $a$
equal to $2 a$
less than $2 a$
greater than $2 a$

Toppr Admin · Accepted Answer

F-x2212-f-ma1a1-a which is givenF-f-ma1Now applied force is doubled therefore2F-x2212-f-ma2ma2-2-f-ma1-x2212-fThis gives ma2-f-ma1-ma1-f-2ma1Therefore- clearly- a2-gt-2a1a2-gt-2a

A body moves on a rough horizontal surface with acceleration a when it is subjected to a horizontal force. If the applied force is doubled, it moves with accelerationequal to aequal to 2aless than 2agreater than 2a

F−f=ma1a1=a which is givenF=f+ma1Now applied force is doubled therefore2F−f=ma2ma2=2(f+ma1)−fThis gives ma2=f+ma1+ma1=f+2ma1Therefore, clearly, a2>2a1a2>2a

A body moves on a rough horizontal surface with acceleration $a$ when it is subjected to a horizontal force. If the applied force is doubled, it moves with acceleration
equal to $a$
equal to $2 a$
less than $2 a$
greater than $2 a$

$F - f = m a_{1}$
$a_{1} = a$ which is given
$F = f + m a_{1}$
Now applied force is doubled therefore
$2 F - f = m a_{2}$
$m a_{2} = 2 (f + m a_{1}) - f$
This gives $m a_{2} = f + m a_{1} + m a_{1} = f + 2 m a_{1}$
Therefore, clearly, $a_{2} > 2 a_{1}$
$a_{2} > 2 a$