Two smooth blocks A of mass 1 kg and B of mass 2 kg are connected by alight string passing over a smooth pulley as shown….

Q: Two smooth blocks A of mass 1 kg and B of mass 2 kg are connected by alight string passing over a smooth pulley as shown. The block B is sliding down with a velocity 2 m/s. A force F is applied on the block A so that the block B will reverse its direction of motion after 3 s.
Numerical

(i) Find the acceleration of block A.
(a) 1 m/s²
(b) 2/3 m/s²
(c) 3 m/s²
(d) 4/9 m/s²

Ans: (b)

(ii) Find the tension in the string.
(a) 10 N
(b) 11.33 N
(c) 13.67 N
(d) 40 N

Ans: (b)

Sol:(i) When direction of motion will reverse at the instant, the velocity becomes zero.
∵ v = u + at

Or 0 = 2 + a×3

a = -2/3 m/s2

Thus, acceleration of A is 2/3 m/s2 in rightward direction.

Sol:(ii)
From FBD of A,

F -T = m1 a = 1 × 2/3 ….(i)

From FBD of B, ….(ii)

T – m2 g sinθ = m2 a

= 2 × 2/3 = 4/3

or,  T-10 = 4/3

T= 10 + 4/3 = 34/3

=11.33 N

A block of mass m is to be raised by a light rope from rest to rest through a height 60m, the maximum tension….

Q: A block of mass m is to be raised by a light rope from rest to rest through a height 60m, the maximum tension of which the rope safely bear is 4 mg. The least time in which the ascent can be made is

(a) 1 s

(b) 2 s

(c) 3 s

(d) 4 s

Ans: (d)
Sol: Let up to t = t1 block is accelerated with maximum acceleration , then decelerated with deceleration g .
\displaystyle h_1 = \frac{1}{2}(\frac{4mg-mg}{m})t_1^2

vo = 3gt1
Also
0 = vo – g(t-t1)
t1 = t/4 ,
h2 = vo2/2g
h = h1 + h2
On solving
t= 4sec

For the situation shown in figure, all pulleys are smooth and fixed. The inclined plane is also smooth….

Q: For the situation shown in figure, all pulleys are smooth and fixed. The inclined plane is also smooth. A constant force F of 38 N is applied at the end of the string. Find the ratio of magnitudes of tensions in the string A and B.

Numerical

(a) 23 : 57

(b) 32 : 57

(c) 23 : 75

(d) 32 : 75

Ans: (a)

Sol:
Let tension in the string between 1kg and 2kg block is T1 & in other part is T2

For 1 kg block ,

T1 – 1×g = 1×a   ..(i)

For 2kg block ;

T2 – 2g sin37° -T1 = 2a   …(ii)

since T2 = F = 38 N

from (ii)

38 – 2×10 ×(3/5) – (g+a) = 2a

(As , T1 = g+a & sin37° = 3/5)

38 – 12 -10 = 3a

16 = 3a

a = 16/3 m/s2

Since , T1 = g+a

T1 = 10 + 16/3 = 46/3 N

\displaystyle \frac{T_1}{T_2} = \frac{46/3}{38}= \frac{23}{57}