# How is it done(physics)?

## Question; 1)The magnitude of acceleration of the 10 kg block is : 2)The magnitude of acceleration of the 15 kg block is : 3)If applied force F = 120 N, then magnitude of acceleration of 15 kg block will be 4) 5)In the situation of the previous question, acceleration of the 15 kg block will be :

Apr 9, 2016

(1) $2 m {s}^{-} 2$. (2) $4 m {s}^{-} 2$. (3) $4 m {s}^{-} 2$. (4) $2 m {s}^{-} 2$, towards right
(5) $4 m {s}^{-} 2$, towards right

#### Explanation:

(1). Normal reaction $N$ of the $10 k g$ block $= m g$
$= 10 \times 10 = 100 N$

Frictional force ${F}_{f} = \mu \times N = 0.6 \times 100 = - 60 N$,
$- v e$ sign indicates that frictional force is in opposite direction of the applied force.
Net force${F}_{\text{net"=F_"applied}} + {F}_{f}$
$= 80 - 60 = 20 N$
Acceleration of $10 k g$ block can be found from the expression $F = m \cdot a$
or ${a}_{10} = \frac{F}{m} = \frac{20}{10} = 2 m {s}^{-} 2$

(2). Let us examine what is the force acting on $15 k g$ Block by $10 k g$ block. Due to friction, there is interlock of cavities/crevices existing on the faces of contact. The force of friction which opposes motion of $10 k g$ block will create reaction force on $15 k g$.

We know that action and reaction are equal and opposite. Hence, there will be a reaction of $60 N$ on $15 k g$ block. As the other end of this block rests on a friction-less surface, the acceleration produced in it is
${a}_{15} = \frac{60}{15} = 4 m {s}^{-} 2$

(3). For an applied force of $120 N$
There is no change of force of friction. Hence, there is no change in the reaction force on $15 k g$ Block which remains as $= 60 N$.

$\therefore {a}_{15} = \frac{60}{15} = 4 m {s}^{-} 2$

(4). Assuming that pulleys are friction-less, these are used to change the direction of applied force. The magnitude remains the same.

$\therefore$ Acceleration of $10 k g$ Block is same as it was in part (a) of the question.

(5). By the same logic applied as in (4). above

${a}_{15} = 4 m {s}^{-} 2$ towards right.