Q#7
Two blocks A and B of mass mA and mB respectively are kept in contact on a friction-less table. The experimenter pushes the block A from behind so that the blocks accelerate. If the block A exerts a force F on the block B, what is the force exerted by the experimenter on A?Answer:
Both the blocks accelerate simultaneously with same acceleration, say a. Since block A exerts a force F on block B which mass is mB . So, a = F/mB. Now the force exerted by the experimenter (say P) accelerates both the blocks combined mass of which is mA + mB. So,
P = (mA + mB)F/mB = F(1 + mA/mB).
Q#8
Raindrops of radius 1 mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head.
Answer:
First of all we need to know the acceleration 'a' which is negative here. The data given:distance s = 1 mm = 1 x 10-3 m, initial velocity u = 30 m/s, final velocity v = 0 , we use the equation
v² = u² + 2as
0² = 30² + 2a(1 x 10-3m)
a = -900/2 x 10-3 = -450000 m/s² (Negative sign indicates that it is a retardation)
mass m = 4 mg = 4 x 10-6 kg
So, the force exerted by each drop on the head = ma
= (4 x 10-6 kg)(450000 m/s²) = 1.80 N.
Q#9
A particle of mass 0.3 kg is subjected to a force F = -kx with k = 15 N/m. What will be the initial acceleration if it is released from a point x = 20 cm?
Answer:
From the given data we have force F = (15 N/m)(0.20 m)
= 3 N, mass m = 0.3 kg. From the relation F = ma, we get the initial acceleration
a = F/m = 3/0.3 m/s² = 10 m/s².
Q#10
Both the springs shown in figure (5-E2) are unstretched. If the block is displaced by a distance x and released, what will be the initial acceleration?
We know the mass of the particle = m. To know the initial acceleration we need to know the net force acting on the block due to the springs.
Since at the original position the springs are unstretched, so whether we displace the block to the left or to the right, both the springs will apply force along the same direction ie towards original position opposing the displacement. Hence both the forces by the springs will add and its value
= k1x + k2x = (k1 + k2)x
So at this position the initial acceleration of the block will be
= Force/mass = (k1 + k2)x/m and its direction will be opposite to the displacement.
Q#11
A small block B is placed on another block A of mass 5 kg and length 20 cm. Initially the block B is near the right end of block A (figure 5-E3). A constant horizontal force of 10 N is applied to the block A. All the surfaces are assumed friction-less. Find the time elapsed before the block B separates from A.
Answer:
Since all the surfaces are friction-less so block B will not move but the block A will slide under it. When the block A slides its full length 20 cm, the block B will separate from it.
Mass m of block A = 5 kg, Applied force F = 10 N ,so acceleration a = F/m
= 10/5 m/s² = 2 m/s²
For the block A initial velocity u = 0, acceleration a = 2m/s², distance to be traveled s = 20 cm = 0.20 m, we need to know the time t = ? Here we will use the relation
s = ut + ½at²
0.20 = 0 + ½(2)t²
t² = 0.2
t = 0.45 s.
Q#12
A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure (5-E4). Show that the force (assumed equal for both the friends) exerted by each friend on the rope increases as the man moves up . Find the force when the man is at a depth h.
Answer:
Let the mass of the fallen man be m. His weight = mg (downwards)
Let the force exerted by the friends = F = tension in the rope
If the rope makes an angle θ with the vertical near the man, the component of tension in the rope that helps him to lift upwards
= Fcos θ; for both the ropes = 2F cos θ
To lift the man it should be more than his weight, At balancing position,
2F cos θ = mg
F = mg/2cos θ (i)
It is clear from this equation that the force F is inversely proportional to cos θ . As the man comes up the angle between the rope and vertical = θ increases because the rope tends to be horizontal. But as the θ increases the value of cos θ decreases. And as cos θ decreases the value of F increases. So, the force exerted by the friends increases as the man moves up.
From the figure cos θ = h/√{h² + (d/2)²} , putting this value in (i) we get,
F = mg{√h² + (d/2)²}/2h = mg√(4h² + d²)/4h
It is the value of force exerted by each friend at depth h.
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