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WORK AND ENERGY:--
EXERCISES (21-30)
Answer: Potential Energy of the projectile with respect to the ground when it is projected = mgh = mg*40 J
22. The 200 m freestyle women's swimming gold medal at Seol Olympic 1988 went to Heike Friendrich of East Germany when she set a new Olympic record of 1 minute and 57.56 seconds. Assume that she covered most of the distance with a uniform speed and had to exert 460 W to maintain her speed. Calculate the average force of resistance offered by the water during the swim.
23. The US athlete Florence Griffith Joyner won the 100 m sprint gold medal at Seol Olympic 1988 setting a new Olympic record of 10.54 s. Assume that she achieved her maximum speed in a very short time and then ran the race with that speed till she crossed the line. Take her mass to be 50 kg. (a) Calculate the kinetic energy of Griffith Joyner at her full speed.(b) Assume that the track, the wind etc. offered an average resistance of 1/10th of her weight, calculate the work done by the resistance during the run.(c) What power Griffith Joyner had to exert to maintain uniform speed?
→a=(M-m)g/(M+m) =(3.0-2.0)g/(3.0+2.0) =g/5
EXERCISES (21-30)
21. A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground.
Answer: Potential Energy of the projectile with respect to the ground when it is projected = mgh = mg*40 J
Kinetic Energy when projected = ½mv² =½m(50)² = m*1250 J.
Total energy at the time of projection = K.E+P.E. = (mg*40+m*1250) J
Let the speed of the projectile when it hits the ground = V, then at this point its P.E. = 0 (Since h=0)
K.E. = ½mV² J
So, total energy = K.E.+P.E. =½mV² J
Total Energy will remain constant.
So, ½mV²= mg*40+m*1250
V² = 80g+2500 = 80*9.8+2500 = 3284
V = √3284 = 57.31 m/s
22. The 200 m freestyle women's swimming gold medal at Seol Olympic 1988 went to Heike Friendrich of East Germany when she set a new Olympic record of 1 minute and 57.56 seconds. Assume that she covered most of the distance with a uniform speed and had to exert 460 W to maintain her speed. Calculate the average force of resistance offered by the water during the swim.
Answer: Let the force applied by swimmer = F and force of resistance offered by the water = R
So resultant force in the direction of motion = F-R
Since the swimmer swims with a uniform speed, resultant force is zero, i.e. F-R=0, →F=R
Since she exerted 460 W of power, so water resistance also resisted with 460 W of power.
If v= uniform speed of the swimmer =200 m/(60+57.56) s =1.70 m/s
Now Power =Force x speed
→460 = R x 1.70
→R= 460/1.70 = 270.60 N
Since the swimmer swims with a uniform speed, resultant force is zero, i.e. F-R=0, →F=R
Since she exerted 460 W of power, so water resistance also resisted with 460 W of power.
If v= uniform speed of the swimmer =200 m/(60+57.56) s =1.70 m/s
Now Power =Force x speed
→460 = R x 1.70
→R= 460/1.70 = 270.60 N
23. The US athlete Florence Griffith Joyner won the 100 m sprint gold medal at Seol Olympic 1988 setting a new Olympic record of 10.54 s. Assume that she achieved her maximum speed in a very short time and then ran the race with that speed till she crossed the line. Take her mass to be 50 kg. (a) Calculate the kinetic energy of Griffith Joyner at her full speed.(b) Assume that the track, the wind etc. offered an average resistance of 1/10th of her weight, calculate the work done by the resistance during the run.(c) What power Griffith Joyner had to exert to maintain uniform speed?
Answer: Uniform speed of the Athlete v= 100 m/10.54 s =9.49 m/s
Mass m = 50 kg
(a) Kinetic Energy of the Athlete at full speed =½mv² =0.5x50x(9.49)² = 2251 J
(b) Weight of the Athlete = mg =50 x 9.8 N =490 N
Force of resistance =1/10 of 490 N =49 N
Since she runs with a constant speed, the Net force on her in the direction of movement is zero, i.e. Force exerted by her and the force of resistance are equal in magnitude and opposite in direction. So work done by the resistance = Force x distance
= 49 N x -100 m = -4900 J (Negative sign is for the movement opposite to the Force)
(c) Since force applied by the Athlete and the resistance are equal in magnitude but opposite in direction, So force applied F = 49 N,
Constant speed of the Athlete =v = 9.49 m/s
So power exerted by her = F x v =49 x9.49 W =465 W
Mass m = 50 kg
(a) Kinetic Energy of the Athlete at full speed =½mv² =0.5x50x(9.49)² = 2251 J
(b) Weight of the Athlete = mg =50 x 9.8 N =490 N
Force of resistance =1/10 of 490 N =49 N
Since she runs with a constant speed, the Net force on her in the direction of movement is zero, i.e. Force exerted by her and the force of resistance are equal in magnitude and opposite in direction. So work done by the resistance = Force x distance
= 49 N x -100 m = -4900 J (Negative sign is for the movement opposite to the Force)
(c) Since force applied by the Athlete and the resistance are equal in magnitude but opposite in direction, So force applied F = 49 N,
Constant speed of the Athlete =v = 9.49 m/s
So power exerted by her = F x v =49 x9.49 W =465 W
24. A water pump lifts water from a level 10 m below the ground. Water is pumped at a rate of 30 kg/minute with negligible velocity. Calculate the minimum horsepower the engine should have to do this.
Answer: Since pumped water has negligible velocity, its K.E. is zero. The pump has to do work in increasing the P.E. of water to 10 m above.
Water pumped per second = 30 kg/60 s =0.5 kg/s
Weight of this water =0.5 x 9.8 N = 4.9 N
So the pump has to apply 4.9 N force against the gravity to lift it. Hence work done by the pump =4.9 Nx 10 m =49 J/s =49 W
So minimum power of the engine = 49 W =49/746 hp =0.066 hp =6.6x10-2 hp
Water pumped per second = 30 kg/60 s =0.5 kg/s
Weight of this water =0.5 x 9.8 N = 4.9 N
So the pump has to apply 4.9 N force against the gravity to lift it. Hence work done by the pump =4.9 Nx 10 m =49 J/s =49 W
So minimum power of the engine = 49 W =49/746 hp =0.066 hp =6.6x10-2 hp
25. An unruly demonstrator lifts a stone of mass 200 g from the ground and throws it at his opponent. At the time of projection, the stone is 150 cm above the ground and has a speed of 3.00 m/s. Calculate the work done by the demonstrator during the process. If it takes 1 second for the demonstrator to lift the stone and throw, what horsepower does he use?
Answer: Mass of the stone m = 200 g =0.20 kg
Height from the ground h= 150 cm =1.50 m
So work done against gravity in lifting the stone = mgh =0.20x9.8x1.50 = 2.94 J
Speed given to the stone v= 3.00 m/s
K.E. =½mv² =0.50x0.20x3² = 0.90 J
So work done in increasing the K.E. of the stone =0.90 J
Hence Total work done by the demonstrator= 2.94 J+0.90 J =3.84 J
Time taken in this process = 1 s
So power used by the demonstrator = 3.84 J/1 s =3.84 W =3.84/746 hp = 5.15x10-3 hp
26. In a factory, it is desired to lift 2000 kg of metal through a distance of 12 m in 1 minute. Find the minimum horsepower of the engine to be used.A
Answer: Minimum work is done by the engine when the metal is lifted slowly. Mass of the metal = 2000 kg, Distance of the lift =12 m.
So work done against the gravity =mgh = 2000x9.8x12 J =235200 J
Time taken =1 m =60 s =60 s
Minimum horsepower required =235200/60 W =3920 W =3920/746 hp = 5.25 hp
So work done against the gravity =mgh = 2000x9.8x12 J =235200 J
Time taken =1 m =60 s =60 s
Minimum horsepower required =235200/60 W =3920 W =3920/746 hp = 5.25 hp
27.A scooter company gives the following specifications about its product
Weight of the scooter - 95 kg
Maximum speed - 60 km/h
Maximum engine power - 3.5 hp
Pickup time to get the maximum speed - 5 s
Check the validity of these specifications.
Answer: The engine makes the scooter to change the speed from zero to v=60 km/h =60000/3600 m/s =16.67 m/s
K.E. of the scooter at rest =0
K.E. of the scooter at maximum speed =½mv² =0.5x95x(16.67)² =13194.44 J
Work done by the engine = Change in K.E. =13194.44 J
Time taken =5 s
Required Power of the engine = 12194.44/5 =2638.88 W =2638.88/746 hp =3.54 hp
But the quoted power =3.5 hp
So with 3.5 hp power of the engine, the scooter cannot achieve exactly 60 km/h speed in 5 s. It is somewhat overclaimed.
K.E. of the scooter at rest =0
K.E. of the scooter at maximum speed =½mv² =0.5x95x(16.67)² =13194.44 J
Work done by the engine = Change in K.E. =13194.44 J
Time taken =5 s
Required Power of the engine = 12194.44/5 =2638.88 W =2638.88/746 hp =3.54 hp
But the quoted power =3.5 hp
So with 3.5 hp power of the engine, the scooter cannot achieve exactly 60 km/h speed in 5 s. It is somewhat overclaimed.
28. A block of mass 30.0 kg is being brought down by a chain. If the block acquires a speed of 40.0 cm/s in dropping down 2.00 m, find the work done by the chain during the process.
Answer: Let force applied by the chain = T Newton (Upward)
Weight of the block = 30.0 kg x 9.8 m/s² =294 N (Downward)
Final speed v = 40.0 cm/s =0.40 m/s
Change in K.E. =½*30*(0.40)² =2.4 J
So work done by all the forces on the block =2.4 J
Forces on the block are force by chain and force of gravity i.e. Net force (downward) = 294-T N
Work done by this force in dropping the block by 2.00 m
(294-T)x2.00 = 2.4
→294-T=1.2
→T=294-1.2 =292.8 N
The distance travelled by the block is opposite to the direction of the force applied by the chain = -2.00 m
So work done by the chain during the process = 292.8 N x (-2.00 m) =-585.6 J ≈-586 J
Weight of the block = 30.0 kg x 9.8 m/s² =294 N (Downward)
Final speed v = 40.0 cm/s =0.40 m/s
Change in K.E. =½*30*(0.40)² =2.4 J
So work done by all the forces on the block =2.4 J
Forces on the block are force by chain and force of gravity i.e. Net force (downward) = 294-T N
Work done by this force in dropping the block by 2.00 m
(294-T)x2.00 = 2.4
→294-T=1.2
→T=294-1.2 =292.8 N
The distance travelled by the block is opposite to the direction of the force applied by the chain = -2.00 m
So work done by the chain during the process = 292.8 N x (-2.00 m) =-585.6 J ≈-586 J
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) .................. (ii)
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+½x(g/3)x1² = g/6
 |
| Figure for Q-29 |
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) .................. (ii)
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+½x(g/3)x1² = 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)xgxg/6 -(3T/4g)xgxg/6 =Tg/4-Tg/8 = Tg/8 = 16x9.8/8 = 19.6 J
→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)xgxg/6 -(3T/4g)xgxg/6 =Tg/4-Tg/8 = Tg/8 = 16x9.8/8 = 19.6 J
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 ------(i)
and
Mg-T= Ma → T=Mg-Ma -----(ii)
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]
=½a{t²-(t-1)²}
=½x(g/5)x{4²-3²}
=½x(g/5)x7
=7g/10 m
Hence work done by the gravity,
=Mgs-mgs
=(M-m)gs
=(3-2)gx(7g/10)
=7g²/10
=7x9.8²/10
=67.23 J
s = (ut+½at²)-{u(t-1)+½a(t-1)²} [here t=4 s, u=0]
=½a{t²-(t-1)²}
=½x(g/5)x{4²-3²}
=½x(g/5)x7
=7g/10 m
Hence work done by the gravity,
=Mgs-mgs
=(M-m)gs
=(3-2)gx(7g/10)
=7g²/10
=7x9.8²/10
=67.23 J
===<<<O>>>===
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CHAPTER- 10 - Rotational Mechanics
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