According to the potential energy-displacement graph in Fig. 7.39, total mechanical energy is fixed at 30 J. At point P, the graph indicates potential energy is 20 J, leaving 10 J for kinetic energy, which gives a speed of 6.32 m s-1. At point Q, potential energy is 30 J, meaning kinetic energy is 0Read more
According to the potential energy-displacement graph in Fig. 7.39, total mechanical energy is fixed at 30 J. At point P, the graph indicates potential energy is 20 J, leaving 10 J for kinetic energy, which gives a speed of 6.32 m s-1. At point Q, potential energy is 30 J, meaning kinetic energy is 0 J and velocity is 0 m s-1. The ball cannot reach R.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
Using the speed-time graph in Fig. 7.38, the car moves with uniform speed at 35 m s-1 from time 0 to 1 second between A and B. Its kinetic energy at A is half multiplied by 1000 kg multiplied by 35 squared, equaling 612500 J. Between B and C, the brakes bring speed to zero, doing minus 612500 J of wRead more
Using the speed-time graph in Fig. 7.38, the car moves with uniform speed at 35 m s-1 from time 0 to 1 second between A and B. Its kinetic energy at A is half multiplied by 1000 kg multiplied by 35 squared, equaling 612500 J. Between B and C, the brakes bring speed to zero, doing minus 612500 J of work, dissipating energy as heat and sound.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
The maximum height attained by an object thrown vertically upward is given by initial velocity squared divided by twice the gravitational acceleration. With the same initial velocity, height is inversely proportional to gravity. Because lunar gravity is one-sixth of terrestrial gravity, the height rRead more
The maximum height attained by an object thrown vertically upward is given by initial velocity squared divided by twice the gravitational acceleration. With the same initial velocity, height is inversely proportional to gravity. Because lunar gravity is one-sixth of terrestrial gravity, the height reached on the Moon is six times the height on Earth, calculating to six times 8 m, which equals 48 m.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
From Fig. 7.37, initial kinetic energy of 180 J yields an initial speed of 6 m s-1. The work done equals the area under the force-displacement graph, calculated as half times the sum of parallel sides 2 m and 4 m multiplied by height 50 N, giving 150 J. Final kinetic energy becomes 330 J, giving a sRead more
From Fig. 7.37, initial kinetic energy of 180 J yields an initial speed of 6 m s-1. The work done equals the area under the force-displacement graph, calculated as half times the sum of parallel sides 2 m and 4 m multiplied by height 50 N, giving 150 J. Final kinetic energy becomes 330 J, giving a speed of 8.12 m s-1 at 4 m. Acceleration remains non-negative throughout.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
During upward motion, gravity opposes displacement giving negative work, while downward gravity aligns with displacement giving positive work. Initial kinetic energy is half mass velocity squared, equaling 400 J. Gravitational work to 19.4 m equals minus m g h, which is minus 388 J. By the work-enerRead more
During upward motion, gravity opposes displacement giving negative work, while downward gravity aligns with displacement giving positive work. Initial kinetic energy is half mass velocity squared, equaling 400 J. Gravitational work to 19.4 m equals minus m g h, which is minus 388 J. By the work-energy theorem, net work equals change in kinetic energy, revealing that air resistance does minus 12 J of work.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
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.
According to the potential energy-displacement graph in Fig. 7.39, total mechanical energy is fixed at 30 J. At point P, the graph indicates potential energy is 20 J, leaving 10 J for kinetic energy, which gives a speed of 6.32 m s-1. At point Q, potential energy is 30 J, meaning kinetic energy is 0Read more
According to the potential energy-displacement graph in Fig. 7.39, total mechanical energy is fixed at 30 J. At point P, the graph indicates potential energy is 20 J, leaving 10 J for kinetic energy, which gives a speed of 6.32 m s-1. At point Q, potential energy is 30 J, meaning kinetic energy is 0 J and velocity is 0 m s-1. The ball cannot reach R.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-7/
See lessA 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. (ii) Calculate the kinetic energy of the car at A. (iii) State the work done by the brakes in bringing the car to a halt between B and C. (iv) What does the kinetic energy of the car transform into?
Using the speed-time graph in Fig. 7.38, the car moves with uniform speed at 35 m s-1 from time 0 to 1 second between A and B. Its kinetic energy at A is half multiplied by 1000 kg multiplied by 35 squared, equaling 612500 J. Between B and C, the brakes bring speed to zero, doing minus 612500 J of wRead more
Using the speed-time graph in Fig. 7.38, the car moves with uniform speed at 35 m s-1 from time 0 to 1 second between A and B. Its kinetic energy at A is half multiplied by 1000 kg multiplied by 35 squared, equaling 612500 J. Between B and C, the brakes bring speed to zero, doing minus 612500 J of work, dissipating energy as heat and sound.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-7/
See lessThe 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?
The maximum height attained by an object thrown vertically upward is given by initial velocity squared divided by twice the gravitational acceleration. With the same initial velocity, height is inversely proportional to gravity. Because lunar gravity is one-sixth of terrestrial gravity, the height rRead more
The maximum height attained by an object thrown vertically upward is given by initial velocity squared divided by twice the gravitational acceleration. With the same initial velocity, height is inversely proportional to gravity. Because lunar gravity is one-sixth of terrestrial gravity, the height reached on the Moon is six times the height on Earth, calculating to six times 8 m, which equals 48 m.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-7/
See lessA 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 (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?
From Fig. 7.37, initial kinetic energy of 180 J yields an initial speed of 6 m s-1. The work done equals the area under the force-displacement graph, calculated as half times the sum of parallel sides 2 m and 4 m multiplied by height 50 N, giving 150 J. Final kinetic energy becomes 330 J, giving a sRead more
From Fig. 7.37, initial kinetic energy of 180 J yields an initial speed of 6 m s-1. The work done equals the area under the force-displacement graph, calculated as half times the sum of parallel sides 2 m and 4 m multiplied by height 50 N, giving 150 J. Final kinetic energy becomes 330 J, giving a speed of 8.12 m s-1 at 4 m. Acceleration remains non-negative throughout.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-7/
See lessA 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. (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).
During upward motion, gravity opposes displacement giving negative work, while downward gravity aligns with displacement giving positive work. Initial kinetic energy is half mass velocity squared, equaling 400 J. Gravitational work to 19.4 m equals minus m g h, which is minus 388 J. By the work-enerRead more
During upward motion, gravity opposes displacement giving negative work, while downward gravity aligns with displacement giving positive work. Initial kinetic energy is half mass velocity squared, equaling 400 J. Gravitational work to 19.4 m equals minus m g h, which is minus 388 J. By the work-energy theorem, net work equals change in kinetic energy, revealing that air resistance does minus 12 J of work.
For more NCERT Solutions of Class 9 Science Exploration Chapter 7 Work, Energy and Simple Machines Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/science/exploration-chapter-7/
See less