Question

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 :

Answer 1

(1) `2ms^-2`. (2) `4ms^-2`. (3) `4ms^-2`. (4) `2ms^-2`, towards right
(5) `4ms^-2`, towards right

Explanation:

(1). Normal reaction `N` of the `10kg` block `=mg`
`=10xx10=100 N`

Frictional force `F_f=muxxN=0.6xx100=-60N`,
`-ve` sign indicates that frictional force is in opposite direction of the applied force.
Net force`F_"net"=F_"applied"+F_f`
`=80-60=20N`
Acceleration of `10kg` block can be found from the expression `F=mcdot a`
or `a_10=F/m=20/10=2ms^-2`

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

We know that action and reaction are equal and opposite. Hence, there will be a reaction of `60N` on `15kg` block. As the other end of this block rests on a friction-less surface, the acceleration produced in it is
`a_15=60/15=4ms^-2`

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

`:. a_15=60/15=4ms^-2`

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

`:.` Acceleration of `10kg` Block is same as it was in part (a) of the question.

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

`a_15=4ms^-2` towards right.