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WORK AND ENERGY:--
EXERCISES (31-42)
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.
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).
38. A uniform chain of mass m and length l overhangs a table with its two-third part on the table. Find the work to be done by a person to put the hanging part back on the table.
Answer: K.E. given to the block = ½mv²
Let it rise to a height of h,
At the highest point change in P.E. = mgh
At this point total K.E. is converted to P.E.
i.e. mgh = ½*mv²
→h = v²/2g =2x2/(2x10) m =1/5 m =100/5 cm =20 cm
42. A block of mass 250 g is on a vertical spring of spring constant 100 N/m fixed from below.The spring is now compressed to have a length 10 cm shorter than its natural length and the system is released from this position. How high does the block rise? Take g=10 m/s².
Answer: Work done on the spring =½kx² =½*100*(0.10)² =0.50 J
=Potential energy stored in the spring
When the system is released the block rises to a height of, say h. At the highest point P.E. =mgh
It will be equal to P.E. stored in the spring.
So, mgh = 0.50
→h = 0.50/(0.25x10) m = 0.20 m = 20 cm
EXERCISES (31-42)
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.
 |
| Figure for problem 31 |
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*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+39.2µ ------------------- (a)
Acceleration of 1.0 kg block will be half of bigger block. So a'= a/2
Here m'g-2T=m'a'
→1.0*g-2T=1.0*a/2
→g-2T=a/2
→2T=g-a/2
→T=g/2-a/4 -------------------(b)
Equating (a) and (b) we get,
0.36+39.2µ=9.8/2-0.09/4
→39.2µ=4.9-0.36-0.0225 =4.52
→µ=0.12
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
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*15² = 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 = -20,300 J
So work done against the friction = 20,300 J
Initial speed = 54 km/h = 54000/3600 m/s = 15 m/s
Initial K.E. = ½mv² =½*1400*15² = 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 = -20,300 J
So work done against the friction = 20,300 J
34. A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3.2 m high. How much work was required (a) to lift the block from the ground and put it at the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground if (c) it falls off the incline and drops vertically on the ground (d) it slides down the incline? Take g=10 m/s².
Answer: (a) Mass of the block = 200 g = 0.200 kg
Height of incline h=3.2 m, Length of the incline l= 10 m
Required work done to lift the block from the ground and put it at the top = Work done against the gravity = mgh = 0.200*10*3.2 J =6.40 J
(b) Since the incline is frictionless, only force on the block is that of gravity. Even if it is slid on the incline work is done against the gravity. The movement against the gravity, in this case, is also same i.e. 3.2 m. So in this case also work done = 6.40 J
(c) When the block falls vertically downward, its speed near the ground v²=u²+2gh = 0+2*10*3.2 = 64
→v = 8 m/s
(d) When it slides down the incline, let its speed near the ground be v. Since the only force on the block is that of gravity, the W.D. by the gravity = change in its Kinetic Energy.
→mg*h = ½mv²
→gh=v²/2
→v²=2gh
→v = √(2gh) = √(2*10*3.2) = √64 = 8 m/s
Height of incline h=3.2 m, Length of the incline l= 10 m
Required work done to lift the block from the ground and put it at the top = Work done against the gravity = mgh = 0.200*10*3.2 J =6.40 J
(b) Since the incline is frictionless, only force on the block is that of gravity. Even if it is slid on the incline work is done against the gravity. The movement against the gravity, in this case, is also same i.e. 3.2 m. So in this case also work done = 6.40 J
(c) When the block falls vertically downward, its speed near the ground v²=u²+2gh = 0+2*10*3.2 = 64
→v = 8 m/s
(d) When it slides down the incline, let its speed near the ground be v. Since the only force on the block is that of gravity, the W.D. by the gravity = change in its Kinetic Energy.
→mg*h = ½mv²
→gh=v²/2
→v²=2gh
→v = √(2gh) = √(2*10*3.2) = √64 = 8 m/s
35. In a children's park, there is a slide which has a total length of 10 m and a height of 8.0 m (Figure 8-E3). A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. The average friction offered by the slide is three tenth of the weight. Find (a) the work done by the ladder on the boy as he goes up, (b) the work done by the slide on the boy as he comes down. (c) Find the work done by forces inside the body of the boy.
 |
| Figure for Q-35 |
Answer: (a) When the boy goes up the ladder applies a normal force on the boy which is equal to the weight of the boy = mg = 200 N. Since the movement and the force both are in the same direction the work done is positive. The height of the ladder =8 m
Hence the work done=200 N*8 m = 1600 J.
(b) The slide does the work on the boy through friction. Frictional force = 3/10th of the weight = 3/10*200 N = 60 N
Length of the slide = 10 m
So work done = 60 N*(-10)m = -600 J
(Negative sign is due to the movement against the force)
(c) Work done by the forces inside the body of the boy will be zero because inside the body the forces (internal force) will be in equal and opposite pairs. So net internal force =0.
Hence the work done=200 N*8 m = 1600 J.
(b) The slide does the work on the boy through friction. Frictional force = 3/10th of the weight = 3/10*200 N = 60 N
Length of the slide = 10 m
So work done = 60 N*(-10)m = -600 J
(Negative sign is due to the movement against the force)
(c) Work done by the forces inside the body of the boy will be zero because inside the body the forces (internal force) will be in equal and opposite pairs. So net internal force =0.
36. Figure (8-E4) shows a particle sliding on a frictionless track which terminates in a straight horizontal section. If the particle starts slipping from the point A, how far away from the track will the particle hit the ground?
 |
| Figure for Q-36 |
Answer: It is clear from the figure that at the point of termination of the track the particle descends through 1.0-0.5 = 0.5 m
Its initial K.E. is zero. At the point of termination of the track, its K.E. will be due to change in P.E. Let its velocity at this point be v.
K.E. = ½mv²
Change in P.E. = mgh, Equating them we get,
½mv²=mgh
→v²/2 =gh
→v²=2gh
→v=√(2gh) =√(2*9.8*0.5) =√9.8 = 3.13 m/s
Now the particle will travel in a parabolic path, The vertical distance traveled =s =0.50 m, with initial velocity = 0. Let t be the time to reach the ground. Then from s=ut+½gt² we get,
0.5 = 0+0.5*9.8*t²
→t² = 1/9.8 = 0.1020
→t = 0.32 s
In this time the distance traveled = speed x time
= 3.13 x 0.32 = 1.00 m from the end of the track.
Its initial K.E. is zero. At the point of termination of the track, its K.E. will be due to change in P.E. Let its velocity at this point be v.
K.E. = ½mv²
Change in P.E. = mgh, Equating them we get,
½mv²=mgh
→v²/2 =gh
→v²=2gh
→v=√(2gh) =√(2*9.8*0.5) =√9.8 = 3.13 m/s
Now the particle will travel in a parabolic path, The vertical distance traveled =s =0.50 m, with initial velocity = 0. Let t be the time to reach the ground. Then from s=ut+½gt² we get,
0.5 = 0+0.5*9.8*t²
→t² = 1/9.8 = 0.1020
→t = 0.32 s
In this time the distance traveled = speed x time
= 3.13 x 0.32 = 1.00 m from the end of the track.
37. A block weighing 10 N travels down a smooth curved track AB joined to a rough horizontal surface (figure 8-E5). The rough surface has a friction coefficient of 0.20 with the block. If the block starts slipping on the track from a point 1.0 m above the horizontal surface, how far will it move on the rough surface?
 |
| Figure for Q-37 |
Answer: Potential energy of the block at Point A with respect to ground = mgh = 10 N x 1.0 m = 10 J
K.E. at A = 0
Total Mechanical Energy at A = P.E.+K.E, = 10 J
P.E. at B = 0
K.E. at B = ½mv² (v= speed at B and mass m =10/g kg)
Total Mechanical Energy at B = ½mv² = ½*(10/g)*v² =5v²/g
Total Mechanical energy will be constant. So,
½*(10/g)*v² = 10
→v² =2g
Now on the rough surface, the force of friction will oppose the motion. Frictional force F = µN, Where N is the normal force on the block which is equal to the weight. N=10 Newton and given that µ = 0.20
So, F = 0.20 x 10 = 2 N
Let us assume that it moves a distance d on the rough track. Then work done by the friction force = Fd = 2(-d) J = -2d J
Since only friction force is acting on the block along the motion, change in K.E. of the block = work done by the friction.
Change in K.E. = 0-½mv² = -½mv²
Equating them we get,
-2d = -½mv²
→d = mv²/4 = (10/g)*2g/4 = 5.0 m
K.E. at A = 0
Total Mechanical Energy at A = P.E.+K.E, = 10 J
P.E. at B = 0
K.E. at B = ½mv² (v= speed at B and mass m =10/g kg)
Total Mechanical Energy at B = ½mv² = ½*(10/g)*v² =5v²/g
Total Mechanical energy will be constant. So,
½*(10/g)*v² = 10
→v² =2g
Now on the rough surface, the force of friction will oppose the motion. Frictional force F = µN, Where N is the normal force on the block which is equal to the weight. N=10 Newton and given that µ = 0.20
So, F = 0.20 x 10 = 2 N
Let us assume that it moves a distance d on the rough track. Then work done by the friction force = Fd = 2(-d) J = -2d J
Since only friction force is acting on the block along the motion, change in K.E. of the block = work done by the friction.
Change in K.E. = 0-½mv² = -½mv²
Equating them we get,
-2d = -½mv²
→d = mv²/4 = (10/g)*2g/4 = 5.0 m
38. A uniform chain of mass m and length l overhangs a table with its two-third part on the table. Find the work to be done by a person to put the hanging part back on the table.
Answer: Unit mass of the chain = m/l
Mass of hanging portion = (l/3)*m/l = m/3
C.G. of the hanging part below the table = ½*l/3 = l/6
So work to be done by the person = (m/3)*g*l/6 = mgl/18
Mass of hanging portion = (l/3)*m/l = m/3
C.G. of the hanging part below the table = ½*l/3 = l/6
So work to be done by the person = (m/3)*g*l/6 = mgl/18
39. A uniform chain of length L and mass M overhangs a horizontal table with its two-third part on the table. The friction coefficient between the table and the chain is µ. Find the work done by the friction during the period the chain slips off the table.
Answer: Weight of the chain on the table = 2Mg/3
Normal force on the chain = Weight of the chain = 2Mg/3
Force of friction on the chain =µN =2µMg/3
As the chain slips on the table its weight on the table and hence normal force and friction force starts decreasing. Just when the chain slips off the friction force becomes zero. Hence average friction force = F =(2µMg/3)/2 = 2µMg/6
Work done by the friction force = Force x distance
= (2µMg/6) x (-2L/3)
= -2µMgL/9
Normal force on the chain = Weight of the chain = 2Mg/3
Force of friction on the chain =µN =2µMg/3
As the chain slips on the table its weight on the table and hence normal force and friction force starts decreasing. Just when the chain slips off the friction force becomes zero. Hence average friction force = F =(2µMg/3)/2 = 2µMg/6
Work done by the friction force = Force x distance
= (2µMg/6) x (-2L/3)
= -2µMgL/9
40. A block of mass 1 kg is placed at a point A of a rough track shown in the figure (8-E6). If slightly pushed towards the right, it stops at point B of the track. Calculate the work done by the frictional force on the block during its transit from A to B.
 |
| Figure for Q-40 |
Answer: The block descends by 1.0 m - 0.8 m = 0.2 m in coming from A to B. Potential energy lost = mgh =1.0 x 10 x 0.2 J = 2.0 J
If there would not have been the frictional force, this lost P.E. would be converted to K.E. Since at point B the block stops, hence no K.E. at B. So, in this case, the negative work done by the frictional force is just equal to the P.E. lost i.e. = -2.0 J
If there would not have been the frictional force, this lost P.E. would be converted to K.E. Since at point B the block stops, hence no K.E. at B. So, in this case, the negative work done by the frictional force is just equal to the P.E. lost i.e. = -2.0 J
41. A block of mass 5.0 kg is suspended from the end of a vertical spring which is stretched by 10 cm under the load of the block. The block is given a sharp impulse from below so that it acquires an upward speed of 2.0 m/s. How high the will it rise? Take g=10 m/s².
Answer: K.E. given to the block = ½mv²
Let it rise to a height of h,
At the highest point change in P.E. = mgh
At this point total K.E. is converted to P.E.
i.e. mgh = ½*mv²
→h = v²/2g =2x2/(2x10) m =1/5 m =100/5 cm =20 cm
42. A block of mass 250 g is on a vertical spring of spring constant 100 N/m fixed from below.The spring is now compressed to have a length 10 cm shorter than its natural length and the system is released from this position. How high does the block rise? Take g=10 m/s².
Answer: Work done on the spring =½kx² =½*100*(0.10)² =0.50 J
=Potential energy stored in the spring
When the system is released the block rises to a height of, say h. At the highest point P.E. =mgh
It will be equal to P.E. stored in the spring.
So, mgh = 0.50
→h = 0.50/(0.25x10) m = 0.20 m = 20 cm
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great work sir....
ReplyDeletePlease upload for chapter 9 & so on.
thanks
Sir in question 41 why haven't u considered elastic potential energy?
ReplyDeleteSir in question 32 why is the difference between initial and final position equal to diameter of tube
ReplyDeleteDear Student,
DeleteHere diameter of the tube is the diameter of the circular loop and not the diameter of the cross-section of the tube.
Since the loop made of tube is in vertical plane and the block starts from the top and comes to rest at the bottom, hence the difference between initial and final position is equal to diameter of the circular loop.
Sir can u pls explain the meaning of
ReplyDelete(a) part of ques no. 35 exercise
Please see the correction. Thank you.
DeleteSir there is language confusion in ques 35 ,I mean (a) and (c) part especially (kindly revise it from book)
ReplyDeleteI have corrected the answers. I think now it makes sense.
DeleteSir, in 35 (a), in my opinion boy doesn't do any work on ladder because there is no displacement in ladder that's why the answer is zero for 35 (a).
ReplyDelete