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

Answer:

(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.