Q#29
The heavier block in an Atwood machine has a mass twice that of the lighter one. The tension in the string is 16.0 N when the system is set into motion. Find the decrease in the gravitational potential energy during the first second after the system is released from rest.Answer:
Let us draw the figure as below
Mass M = 2m
Tension T = 16.0 N
Let acceleration of the blocks = a
Then, Mg – T = Ma
T = M(g – a) = 2m(g – a ) (i)
And, T – mg = ma
T = m(g + a)
By equating we get,
2m(g – a) = m(g + a)
2g – 2a = g + a
3a = g
a = g/3
Let us first find the distance travelled in 1st second = s,
Here t = 1 s,
s = ut + ½at² = 0 + ½ x (g/3) x 1² = g/6
To calculate the P.E. we need to know masses of the blocks. From (ii) we have
T = m(g + a)
m = T/(g + a) = T/(g + g/3) = 3T/4g
And M = 2m = 6T/4g
When the system is released, gravitational P.E. of heavier block decreases while that of lighter block increases. So total decrease of gravitational portential energy
= Mgs – mgs
= (6T/4g) x g x g/6 – (3T/4g) x g x g/6
= Tg/4 – Tg/8 = Tg/8 = 16 x (9.8/8) = 19.6 J
Q#30
The two blocks in an Atwood machine has masses 2.0 kg and 3.0 kg. Find the work done by gravity during the fourth second after the system is released from rest.
Answer:
Here M = 3.0 kg, m = 2.0 kg; For the tension T in the string,
T – mg = ma
T= mg + ma
and
Mg – T = Ma
T = Mg – Ma
Equating for T, we get
mg + ma = Mg – Ma
a(M + m) = (M – m)g
a = (M – m)g/(M + m) =(3.0 – 2.0)g/(3.0 + 2.0) = g/5
Distance covered in fourth second,
s = (ut + ½at²) - {u(t – 1) + ½a(t – 1)²} [here t = 4 s, u = 0]
s = ½a{t² – (t – 1)²}
= ½ x (g/5) x {4² – 3²}
= ½ x (g/5) x 7
s = 7g/10 m
Hence work done by the gravity,
= Mgs – mgs
= (M – m)gs
= (3 – 2)g x (7g/10)
= 7g²/10
= 7 x 9.8²/10
= 67.23 J
Q#31
Consider the situation shown in the figure (8-E2). The system is released from rest and the block of mass 1.0 kg is found to have a speed 0.3 m/s after it has descended through a distance of 1 m. Find the coefficient of Kinetic Friction between the block and the table.
Answer:
The speed of 4.0 kg block will be double of the 1.0 kg block and it will also cover twice the distance than that of 1.0 kg block in same time.
So, when 1.0 kg block descends 1 m, the 4.0 kg block will cover 2 m on the table and its speed will be 0.60 m/s.
Now for 4.0 kg block, we have u = 0, v = 0.60 m/s, distance s = 2.0 m. acceleration a =?
v² = u² + 2as
a = v²/2s
a = 0.36/(2 x 2)
a = 0.09 m/s²
Forces on the block are tension T (in the direction of motion) and the force of friction f (opposite to the direction of motion).
F = µmg = µ(4.0)(9.8 N) = 39.2µ N
(Where µ = the coefficient of Kinetic Friction between the block and the table)
Now from Newton's Law of Motion,
T – f =ma
T – 39.2µ = 4.0(0.09)
T = 0.36 + 39
Acceleration of 1.0 kg block will be half of bigger block. So, a '= a/2
Here m'g – 2T = m'a'
1.0g – 2T = 1.0(a/2)
g – 2T = a/2
T = g/2 – a/4
Equating (a) and (b) we get,
0.36 + 39.2µ = 9.8/2 – 0.09/4
39.2µ = 4.52
µ = 0.12
Q#32
A block of 100 g is moved with a speed of 5.0 m/s at the highest point in a closed circular tube of radius 10 cm kept in a vertical plane. The cross-section of the tube is such that the block just fits in it. The block makes several oscillations inside the tube and finally stops at the lowest point. Find the work done by the tube on the block during the process.
Answer:
Mass of the block = 100 g = 0.100 kg, v = 5.0 m/s.
Initial K.E. = ½mv² = 0.5(0.100)(25) = 1.25 J
Final K.E. = 0
Change in K.E. = 0 – 1.25 J = –1.25J
Difference between original position and final position = diameter of the tube = 2(0.10 m) = 0.20 m
Work done by the gravity = mgh = 0.100(9.8)(0.20) = 0.2 J
Only forces on the block are that of the tube and that of gravity, so the Work done by the gravity-Work done by the tube = Change in K.E.
Wok done by the tube = Work done by the gravity – Change in K.E.
Work done by the tube = 0.2 – (–1.25) J = 1.45 J
Since the work done by the tube is through friction, the movement will always be against the force; so this work done will be negative i.e. = –1.45 J
Q#33
A car weighing 1400 kg is moving at a speed of 54 km/h up a hill when the motor stops. If it is just able to reach the destination which is at the height of 10 m above the point, Calculate the work done against friction (Negative of the work done by the friction).
Answer:
Work done by all the forces on the car = Change in its Kinetic energy
Initial speed = 54 km/h = 54000/3600 m/s = 15 m/s
Initial K.E. = ½mv² = ½(1400)(15)² = 157500 J
Final velocity = 0
Final K.E. = 0
Change in K.E. = 0 – 157500 = –157500 J = W.D. by all the forces.
Work done by the gravity = mgh = 1400(9.8)(–10) = –137200 J
Since only forces on the car are gravity and friction, so
Work done by the friction + Work done by the gravity = –157500 J
Work done by the friction + (–137200 J) = –157500 J
Work done by the friction = –157500 J + 137200 J = –20.3 kJ
So work done against the friction = 20.3 kJ
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