Q#18
Suppose the ceiling in the previous problem is that of an elevator which is going up with an acceleration of 2.0 m/s². Find the elongations.Answer:
When the elevator goes up with an acceleration 'a', downwards pseudo force 'ma' acts on the block.
So total force on the block = mg + ma = m(g+a) = 2 (9.8 + 2) = 2 x 11.8 = 23.6 N
Elongation of the spring = 23.6/k = 23.6/100 m = 0.236 m ≈ 0.24 m
When further 1 kg is added to the block, total mass = 3 kg,
Total force = 3(g + a) = 3 x 11.8 N = 35.4 N
Total elongation = 35.4/k = 35.4/100 m = 0.354 m ≈ 0.36 m,
Hence further elongation = 0.36 - 0.24 = 0.12 m
Q#19
The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite to the direction of the velocity and is proportional to it. The balloon carries a mass M and is found to fall down near the earth,s surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v ?
Answer:
Let force of air resistance for velocity v = F
When the balloon falls down with constant velocity v, Net force on it is = Mg - B - F (Taking downward direction as positive). Its value should be zero as there is no acceleration.
Mg – B – F = 0
F = Mg – B
When the balloon goes upwards with a constant velocity v (after removal of some mass, let remained mass be m), net force on it is = mg – B + F
It should also be equal to zero, so
mg – B + F = 0
Subtracting (2) from (1) we get
(M – m)g – 2F = 0
M – m = 2F/g, Putting the value of F in it, we get
M – m = 2 (Mg – B)/g = 2 (M – B/g)
It is the value of mass removed.
Q#20
An empty plastic box of mass m is found to accelerate up at the rate of g/6 when placed deep inside water. How much sand should be put inside the box so that it may accelerate down at the rate of g/6.
Answer:
Let the force of buoyancy on the box be B upwards. Weight of the box = mg, Net force on the box = B – mg (upwards) gives the box an acceleration = g/6. This gives the relation
B-mg = mg/6
Now let sand of mass m1 is added to the box to give it downward acceleration = g/6.
Net force on the box = (m+ m1)g – B. Now force-acceleration relation will be
(m + m1)g – B = (m + m1)g/6 (2)
Adding (A) and (B) we get,
(m + m1)g – mg = mg/6 + (m+ m1)g/6
m1g= 2mg/6 + m1g/6
m1g – m1g/6 = mg/3
5m1g/6 = mg/3
m1 = 2m/5
Q#21
A force F = v x A is exerted on a particle in addition to the force of gravity, where v is the velocity of the particle and A is a constant vector in the horizontal direction. With what minimum speed a particle of mass m be projected so that it continues to move undeflected with a constant velocity?
Answer:
The force vector F is cross product of velocity vector v and constant vector A, so the direction of F will be perpendicular to both v and A. In order to move the particle undeflected, net force on the particle should be zero. It will be only when force of gravity mg is equal and opposite to F. Since gravity force always acts downwards so the direction of force F should be upwards. For it to happen the particle should be projected horizontally with an angle θ to the constant horizontal vector A.
Since the magnitude of F and gravity force are equal, we have
vA sinθ = mg
v = mg/Asinθ,
For the magnitude v to be minimum, sinθ should be maximum. Maximum value of sinθ is 1 for θ = 90°, so minimum speed v of the particle will be equal to mg/A and projected horizontally at right angle to the constant horizontal vector A.
Q#22
In a simple Atwood machine, two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (figure 5-E7) m1= 300g and m2 = 600g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds. (b) Find the the tension in the string. (c) Find the force exerted by the clamp on the pulley.
Answer:
Let the tension in the string be T after the system is released from rest. Let accelerations of the blocks be 'a'. For the first block, T – m1g = m1a
T = m1a + m1g
For the second block, m2g – T = m2a,
m2g – m1a – m1g = m2a (putting the value of T)
(m1 + m2)a = (m2 – m1)g
a = (m2 – m1)g/(m1 + m2 )
(a) Let us calculate the distance 's' travelled after t =2 s of release. from the relation s = ut + ½at², we have
s = 0 + ½ x (m2 – m1)g/(m1 + m2)(2)² = 2(m2 – m1)g/(m1 + m2)
Putting the value m1 = 0.3 kg and m2 = 0.6 kg we get,
S = 2 x 0.3 x 9.8/0.9 m = 6.5 m
(b) Tension in the string T = m1a + m1g
= [m1(m2 – m1)g/(m1 + m2 )] + m1g
= {m1(m2 – m1) + m1(m1 + m2)}g/(m1 + m2)
= 2m1m2g/(m1 + m2) = 2 x 0.3 x 0.6 x 9.8/0.9 = 3.9 N
(c) Force exerted by the clamp on the pulley
= 2T = 2 x 3.9 N = 7.8 N
Q#23
Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment 2.0 s after the system is set into motion. Find the time elapsed before the string is tight again.
Answer:
Acceleration of the blocks, a = (m2 – m1)g/(m1 + m2)
= (0.6 – 0.3) x 9.8/(0.6 + 0.3) = 0.3 x 9.8/0.9 = 3.26 m/s².
When the larger mass is stopped, smaller mass is moving up with a velocity v which can be calculated from the relation,
v = u + at. Here u = 0, t = 2 s and a = 3.26 m/s².
We get,
v = 0 + 3.26 x 2 = 6.52 m/s.
From this instant the string looses its tension and the smaller block moves up under the gravity force with an acceleration -g till it stops. Just the instant it stops the larger mass begins to gain velocity and the string gets tight again. Let the time elapsed before the string is tight again be t'. During this time initial velocity of the smaller block = u' = 6.52 m/s (as calculated above), Final velocity = v' = 0. From the relation
v' = u' – gt'
0 = 6.52 – 9.8t'
t'= 6.52/9.8 = 2/3 seconds.
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