Two blocks are connected by a string passing over a pulley fixed at the edge of a horizontal table shown in figure, then the acceleration of the system is (M2>M1)
Answers
Answer:This may help you
Explanation:
Since string is massless and friction is absent, hence tension in
the string is same every where.
(a)Let acceleration of the pulley be a
p
. For $${
a }_{ p }$$ to be non-zero
T>m
1
g ...(1)
⇒T>2g ...(2)
From (1) and (2),we get
F>2×(2g)⇒F>40N
Therefore, when F=35N
a
P
=0 and hence a
1
=a
2
=0
(b) As mass of the pulley is negligible
F=2T=0
⇒T=F/2 ⇒T=35N
To lift m
2
,
T≥m
2
g T≥50N
Therefore block m
2
will not move
⇒Tm
1
g=m
1
a
1
⇒15=2a
1
⇒a
1
=
2
15
m/s
2
Constraint equation
y
P
+y
P
−y
1
=constant
⇒2y
P
−y
1
=c $$\Rightarrow
2\dfrac { { d }^{ 2 }{ y }_{ p } }{ { dt }^{ 2 } } -\dfrac { { d }^{ 2 }{ y }_{ 1 } }{ { dt }^{ 2 } } =0$$
a
P
=
2
a
1
=
4
15
m/s
2
(c) When F=140 N
T=70 N
⇒T−m
1
g=m
1
a
1
...(1)
⇒70 N−20 N=2×a
1
⇒a
1
=25 m/s
2
T−m
2
g=m
2
a
2
⇒70 N−50 N=5a
2
⇒a
2
=4 m/s
2
Constraint equation
y
p
−y
2
+y
p
−y
1
=c
⇒2y
p
−y
1
−y
2
=c
⇒
dt
2
d
2
y
p
−
dt
2
d
2
y
1
−
dt
2
d
2
y
2
=0
⇒a
p
=
2
a
1
+a
2
=
2
29
m/s
2