Two blocks P and Q of equal masses 1 kg are placed on a rough inclined plane as shown in the figure . Initially the block P is√2 m behind the block Q . While moving down the incline,blocks P and Q experience retarding force√2 N and 3/√2N respectively .
If the two blocks are released simultaneously,then find the time taken by the blocks to come on the same line on the inclined plane as shown in the figure.
Answers
Explanation:
Ans → coefficients of fraction between plane and block A=0.2
coefficients of fraction between plane and block B=0.3
N
A
=mgcos(45
o
)
N
A
=
2
10 m
⇒N=5
2
m
Frictional force acting on block A=F
RA
=N
A
×0.2
⇒
2
10 m
×
10
2
⇒ F
RA
=
2
m
Frictional force acting on block B=F
RB
⇒
2
10 m
×
10
3
⇒ F
RB
=
2
3 m
acceleration of block A & B clown the plane:-
a=gsin45
o
⇒(10/
2
)
a=5
2
a
r
→ resultant acceleration := ma
r
=ma−F
R
ma
r
=m(5
2
)=
2
3 m
ma
r
=ma−
2
ma
r
=5
2
−3/
2
a
r
=4
2
m/s
2
a
r
=7/
2
m/s
2
∗ Distance travelled by A=5 m
∗ Distance travelled by B=(5−
2
)m
for A:-
s=
2
1
at
1
s=
2
1
(4
2
)t
2
t
2
=(s/2
2
)
for B:-
(s−
2
)=
2
1
×
2
7
t
2
(s−
2
)=
2
2
7
t
2
put t
2
=(s/2
2
)
⇒ s=
2
=
2
2
7
(
2
2
s
)
⇒ s=8
2
or 11.31 m
then → t
2
=
2
2
8
2
⇒
t=2 s
→ solution