NCERT Class 9 Science Chapter 7 Work, Energy, and Simple Machines Solutions English Medium As Per New Syllabus. NCERT Class 9 Science Chapter 7 Work, Energy, and Simple Machines Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 9 Science Chapter 7 Work, Energy, and Simple Machines Question Answer Download PDF and select needs one. CBSE Class 9 Science Textual Solutions.
NCERT Class 9 Science Chapter 7 Work, Energy, and Simple Machines
Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 9 Science Chapter 7 Work, Energy, and Simple Machines Solutions. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 9 Science Textbook Solutions English Medium Solutions for All Subject, You can practice these here.
Work, Energy, and Simple Machines
Chapter – 7
| Pause And Ponder |
1. In the previous chapter, a weightlifter is shown holding a barbell steady in her hands (Fig. 6.8). Is she doing any work on the barbell while holding it steady?
Ans: No. Since the barbell does not move, no displacement occurs. Therefore, no work is done on the barbell.
2. Is the work done by friction on the stack of coins that travels on a rough surface (Fig. 6.13c)-positive, negative or zero?
Ans: The work done by friction is negative because it acts opposite to the direction of motion.
| Pause And Ponder |
1. When you pedal a bicycle on a flat road, your muscles supply energy. In what forms does this muscular energy appear as you ride?
Ans: Muscular energy is converted into kinetic energy of the bicycle and rider. Some energy is also converted into heat and sound due to friction.
| Pause And Ponder |
1. Two objects A and B of mass m and 4 m have the same kinetic energy. What is the ratio of the magnitude of velocities of A and B?
Ans:
2. Does the kinetic energy of an object which moves with constant velocity change with its position?
Ans: No. Since the velocity remains constant, the kinetic energy also remains constant.
| Pause And Ponder |
1. Does the potential energy of an object near the surface of the Earth change if it moves with constant velocity in the horizontal direction? What if the object is gradually raised in the vertical direction?
Ans: The potential energy does not change during horizontal motion because the height remains the same. It increases when the object is raised vertically because its height increases.
| Pause And Ponder |
1. For the situation depicted in Fig. 7.19, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.
Ans: Just before the ball reaches the ground, its potential energy is nearly zero, and its kinetic energy becomes mgh. Therefore,
Mechanical Energy = Kinetic Energy + Potential Energy
= mgh + 0
= mgh
Hence, the mechanical energy remains mgh.
2. You may have seen an exhibit like that in Fig. 7.22 in a science park, where a ball is released from the highest point. Describe how the kinetic energy and potential energy change at points A, B and C. Why do subsequent points, such as C, D and E, usually have lower heights compared to the previous ones? Could it have anything to do with the energy lost due to friction?Fig. 7.22: Ball roller coaster in a science park.
Ans: At A, the ball has high potential energy and low kinetic energy. At B, both potential and kinetic energies are present. At C, the ball has high kinetic energy and low potential energy. The later points (C, D, and E) are lower because some mechanical energy is lost due to friction and air resistance.
| Pause And Ponder |
1. Explain why roads on hills are built to wind around in gentle slopes rather than going straight up (Fig. 4.26)?
Ans: Winding roads act as inclined planes, reducing the force required to move vehicles uphill and making the climb easier and safer.
2. To reach a higher floor, we find climbing an inclined ladder easier in comparison to climbing a vertical ladder (Fig. 7.30). Explain why.Fig. 7.30: Climbing ladders.
Ans: An inclined ladder acts as an inclined plane, reducing the force needed to climb, although the distance travelled is greater.
| Pause And Ponder |
1. Why is it easier to open the lid of a can by using a spoon as shown in Fig. 7.35? Fig. 7.35: Opening the lid by using a spoon.
Ans: A spoon acts as a lever, increasing the turning effect (moment) of the applied force. Therefore, less effort is needed to open the lid.
2. Why do you push an object closer to scissors fulcrum when you want to cut an object which is hard?
Ans: Placing the object closer to the fulcrum increases the mechanical advantage, allowing a greater cutting force with less effort.
3. Throughout history, many designs of perpetual machines (using wheels, weights or magnets) have been proposed but none actually work. Why do all real machines eventually slow down and stop? Explain in terms of work and energy.
Ans: All real machines lose some energy as heat and sound due to friction and air resistance. Since energy is continuously lost, the machine cannot keep working forever and eventually slows down and stops.
| Revise, Reflect, Refine |
1. State whether True or False.
(i) Work is said to be done when a force is applied, even if the object does not move.
Ans: False.
(ii) Lifting a bucket vertically upward results in positive work done on the bucket.
Ans: True.
(iii) The SI unit for both work and energy is joule (J).
Ans: True.
(iv) A motionless stretched rubber band has kinetic energy.
Ans: False.
(v) Energy can change from one form to another.
Ans: True.
2. Fill in the blanks.
(i) Work done = ______ × ______ (in the direction of force).
Ans: Force × displacement.
(ii) 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force.
Ans: 1.
(iii) The expression for kinetic energy of a body of mass m and velocity v is ______.
Ans: ½ mv2
(iv) The potential energy of an object of mass m at a small height h from the Earth’s surface is ______.
Ans: mgh.
(v) Power is defined as the ______ at which work is done.
Ans: Rate.
3. When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct?
(i) The force acting on the ball is zero.
(ii) The acceleration of the ball is zero.
(iii) Its kinetic energy is zero.
(iv) Its potential energy is maximum.
Ans: (iii) Its kinetic energy is zero.
(iv) Its potential energy is maximum.
4. For each of the following situations, identify the energy transformation that takes place:
(i) A truck moving uphill.
Ans: Chemical energy (from fuel) → Kinetic energy + Gravitational Potential energy.
(ii) Unwinding of a watch spring.
Ans: Elastic Potential energy (stored in the deformed spring) → Kinetic energy (mechanical energy of the moving watch parts).
(iii) Photosynthesis in green leaves.
Ans: Light energy (from the sun) → Chemical energy (stored in the plant).
(iv) Water flowing from a dam.
Ans: Gravitational Potential energy (stored water) → Kinetic energy (flowing water).
(v) Burning of a matchstick.
Ans: Chemical energy → Thermal (heat) energy + Light energy.
(vi) Explosion of a fire cracker.
Ans: Chemical energy → Thermal energy + Light energy + Sound energy.
(vii) Speaking into a microphone.
Ans: Sound energy → Electrical energy.
(viii) A glowing electric bulb, and.
Ans: Electrical energy → Light energy + Thermal energy.
(ix) A solar panel.
Ans: Light energy → Electrical energy.
5. A student is slowly lifted straight up in an elevator from the ground level to the top floor of a building. Later, the same student climbs the staircase, all the way to the top. Given that the height of the building is h = 72.5 m, acceleration due to gravity is g = 10 m s–2, and student’s mass is m = 50 kg.
(i) Find the gain in the potential energy if the student is lifted straight up to the top.
Ans: Gain in potential energy (lifted straight up) = mgh
= 50 kg×10 ms −2 ×72.5 m= 36,250 J
(ii) Find the gain in the potential energy when the student climbs the stairs to the same top.
Ans: Gain in potential energy (climbing stairs) = mgh
= 50 kg×10 ms −2 ×72.5 m= 36,250 J.
(iii) What do you conclude about the dependence of the potential energy on the path taken?
Ans: The gravitational potential energy of an object depends only on its initial and final vertical heights, and is completely independent of the path taken to reach that height.
6. A crane lifts a mass m to the 10th floor of a building in a certain time. It then raises the same mass to the 20th floor of the same building in double the time. How much more energy and power are required? Assume that the height of all floors is equal.
Ans: Energy: The 20th floor is twice as high as the 10th floor (2h compared to h). Since potential energy is mgh, lifting it twice as high requires exactly double the energy (work done).
Power: Power is the rate of doing work (P=W/t) Since the crane does double the work but takes double the time W/t = 2W/2t the power required remains the same.
7. Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.
Ans: (i) Factors determining energy: The energy (work done) required depends on the mass of the flag (m) and the height of the flagpole (h) because the work done against gravity is calculated as mgh.
(ii) No, raising the flag slowly or quickly does not change the total amount of work done, as the force (weight) and displacement (height) remain identical.
(iii) Change in power requirement: If the speed is doubled, the time taken to raise the flag is halved. Since Power = Work / time, halving the time means the power requirement is doubled.
8. A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.
Ans: Scooter fuel ratio. (Man = 60 kg, Scooter = 100 kg. Day 2 Son = 40 kg)
Day 1 Total Mass (M1): 60 kg + 100 kg = 160 kg
Day 2 Total Mass (M2): 60 kg + 100 kg + 40 kg = 200 kg
Since the scooter reaches the same velocity v, the kinetic energy required on Day 1 is
½ M1 v2 and on Day 2 is ½ M2 v2
Assuming energy transfer happens entirely due to fuel without losses, the fuel used is directly proportional to the kinetic energy gained.
Ratio of fuel used
The ratio of the fuel used is 4:5.
9. On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced.
Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.
Ans: According to the principle of levers, for the seesaw to balance, the product of the load and the load arm must equal the product of the effort and the effort arm (Load × Load Arm = Effort × Effort Arm). Because the adult’s mass is 2 times the child’s mass, the child must sit at twice the distance from the fulcrum compared to the adult to maintain balance (2m×d=m×2d).
10. A ball of mass 2 kg is thrown up with a velocity of 20 m s–1.
(i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion.
Ans: Upward motion → Negative work.
Downward motion → Positive work.
(ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance
(assume g = 10 m s–2).
Ans: Initial K.E.
K.E = ½ mv2
= 1/2 × 2 × (20)2 = 400 J
Potential energy at highest point:
PE=mgh=2×10×19.4=388 J
Work done by air resistance:
Wair = 388 − 400 =−12 J
Wair=−12 J
11. A 10.0 kg block is moving on horizontal floor with negligible friction. As shown in the Fig. 7.37, a variable force is applied on the block in its direction of motion from its position at 0 m till 4 m. If the block had a kinetic energy of 180 J when it was at 0 m, find the block’s speed (i) at 0 m, and.
Ans: Using kinetic energy formula:
KE = 1/2 mv2
180 = 1/2 × 10 × v2
5v2 = 180
5v2 = 180
v2 = 36
v=6 m s−1
(ii) at 4 m. Does the block have negative acceleration in any portion of its motion?
Ans: Work done by the variable force = Area under force–displacement graph
W= ½ × 4 × 50 =100 J
By work-energy theorem:
KEf −180=100
KEf = 280 J
280=½ × 10 × v2 f
5v2f =280
v2 = 56
vf ≈ 7.48 m s−1
Since the force acts in the direction of motion, acceleration is positive.
12. The gravitational attraction on the surface of the Moon (lunar surface) is about 1/6th of that on the surface of the Earth. An astronaut can throw a ball up to a height of 8 m from the surface of the Earth. How far up will the ball thrown with the same upward velocity travel from the surface of the Moon?
Ans: The maximum height reached by a ball thrown upward is:
h = u2/2g
Since the initial velocity u is the same, height is inversely proportional to gravity.
On the Moon:
gMoon = 1/6 gEarth
So, the ball rises 6 times higher on the Moon.
Height on Moon=6×8=48 m
48 m
13. A 1000 kg car is moving along a road at a constant speed. Suddenly, the driver notices some obstruction ahead and applies the brakes to come to a complete stop. The graphical representation of motion of the car starting from the instant the driver spots the traffic ahead is shown in Fig. 7.38.
(i) Describe how the car moves between positions A and B.
Ans: Between A and B, the car moves with a constant speed of 35 m/s. During this interval, the driver is reacting to the obstruction, so the brakes have not yet been applied.
(ii) Calculate the kinetic energy of the car at A.
Ans: Kinetic energy at A: Assuming from typical graphs that the constant speed is 35 m/s,
KE= ½ mv 2
= ½ ×1000 kg× (35) 2
= 500 × 1225
= 612,500 J
(iii) State the work done by the brakes in bringing the car to a halt between B and C.
Ans: The final kinetic energy at C is 0 J.
Using the Work – Energy Theorem:
Work done = Final K.E.−Initial K.E.
= 0 − 612,500
= −612,500 J
(iv) What does the kinetic energy of the car transform into?
Ans: The kinetic energy of the car is transformed mainly into thermal (heat) energy due to friction between the brakes and the wheels, and partly into sound energy.
14. The potential energy-displacement graph of a 0.5 kg ball moving along a frictionless track is shown in Fig. 7.39. At O, the velocity of the ball is 0 m s–1 and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.
Ans: At point O, since velocity is 0, Kinetic Energy = 0. Total Mechanical Energy = KE + PE = 0+30 J=30 J. By the law of conservation of mechanical energy, the total energy remains 30 J at all points along the frictionless track. To find the velocity at points P, Q, and R, you subtract their potential energy (read from the Y-axis of the graph) from 30 J to find the kinetic energy at that point, then use KE=½ mv 2 to find v: v =
(Using estimated graph values where P is at 0 J, Q is at 10 J, and R is at 20 J):
At P: 30 = 0.25v2
At Q: 20 = 0.25v2
At R : 10=0.25v2
15. A coconut of mass 1.5 kg falls from the top of a coconut tree onto the wet sand on a beach. The height of the tree is 10 m. On impact, the coconut comes to rest by making a depression in the sand.
(i) Calculate the velocity of the coconut just before it hits the sand.
Ans: All potential energy converts to kinetic energy mgh= ½ mv 2 ⇒v
≈ 14.14 m/s
(ii) Assume that the average resistive force of sand is 3000 N and all of the coconut’s energy is used to create the depression in the sand. Calculate the depth of the depression the coconut makes in the sand. Assume g = 10 m s–2.
Ans: The total energy of the coconut as it hits the sand is E=mgh=1.5×10×10=150 J
If all this energy is used to do work against the resistive force of the sand to create the depression (Work = Force × distance)
then:
150 J = 3000 N×d
d = 150/3000
=0.05 m
= 5 cm.
Subject Ka Complete Video Solution Dekhein
Step-by-step easy concept explanation aur full questions Answer ke liye watch karein ya app download karein:
Buy Now