How is it done(physics)?

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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
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5)In the situation of the previous question, acceleration of the 15 kg block will be :

1 Answer
Apr 9, 2016

(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 forceF_"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.