Chapter 7: Work, Energy, and Simple Machines
Verified step-by-step solutions for your class notes.
State whether True or False.
- Work is said to be done when a force is applied, even if the object does not move.
- Lifting a bucket vertically upward results in positive work done on the bucket.
- The SI unit for both work and energy is joule (J).
- A motionless stretched rubber band has kinetic energy.
- Energy can change from one form to another.
- False: No movement means zero displacement, so scientifically, no work is done on the object.
- True: The upward lifting force and the upward movement are in the exact same direction.
- True: Energy is just the capacity to do work, so they share the exact same unit.
- False: It's stationary, so its kinetic energy is zero; instead, it stores potential energy.
- True: Energy transfers shapes constantly, like electrical energy turning into light in a bulb.
Fill in the blanks.
- Work done = ______ × ______ (in the direction of force).
- 1 joule of work is done when a force of ______ newton displaces an object by 1 metre in the direction of the force.
- The expression for kinetic energy of a body of mass m and velocity v is ______.
- The potential energy of an object of mass m at a small height h from the Earth’s surface is ______.
- Power is defined as the ______ at which work is done.
- Force × Displacement
- 1
- ½ mv²
- mgh
- rate
When a ball thrown upwards reaches its highest point, tick which of the following statement(s) are correct?
- The force acting on the ball is zero.
- The acceleration of the ball is zero.
- Its kinetic energy is zero.
- Its potential energy is maximum.
The correct statements are: Its kinetic energy is zero. and Its potential energy is maximum.
Even though the ball stalls for an instant at the top (making speed and kinetic energy zero), gravity never turns off! It is still accelerating downwards at 10 m/s² because of its weight. Since it's at the highest peak of its flight, all its energy has converted into maximum potential energy.
For each of the following situations, identify the energy transformation that takes place: a truck moving uphill, unwinding of a watch spring, photosynthesis in green leaves, water flowing from a dam, burning of a matchstick, explosion of a fire cracker, speaking into a microphone, a glowing electric bulb, and a solar panel.
- Truck moving uphill: Chemical energy (fuel) → Kinetic energy → Potential energy.
- Unwinding watch spring:
ElasticPotential energy → Kinetic energy. - Photosynthesis: Light energy (Sun) → Chemical energy.
- Water flowing from a dam: Gravitational Potential energy → Kinetic energy.
- Burning a matchstick: Chemical energy → Thermal (heat) and Light energy.
- Firecracker explosion: Chemical energy → Sound, Light, and Thermal energy.
- Speaking into a mic: Sound energy → Electrical energy.
- Glowing electric bulb: Electrical energy → Light and Thermal energy.
- Solar panel: Light energy → Electrical energy.
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.
- Find the gain in the potential energy if the student is lifted straight up to the top.
- Find the gain in the potential energy when the student climbs the stairs to the same top.
- What do you conclude about the dependence of the potential energy on the path taken?
- Gain = mgh = 50 × 10 × 72.5 = 36,250 J.
- Gain = 36,250 J (The vertical height reached is exactly the same).
- Conclusion: Potential energy is entirely independent of the path taken. It only depends on the starting configuration and final vertical height.
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.
Lifting the mass to the 20th floor means going double the vertical distance, which directly doubles the potential energy required (2 × E). However, because the crane takes double the time to do this double work, the rate of work (Power = Work / Time) stays perfectly identical to the first trip.
It requires twice the energy, but the power required remains exactly the same.
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.
The energy is determined by the mass of the flag and the vertical height of the flagpole. Raising it quickly or slowly does not alter the work done—the mechanical output to combat gravity is always equal to mgh. However, if you double the speed, you finish the task in half the time, which means the power required instantly doubles.
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.
Fuel directly correlates with the kinetic energy (½ mv²) needed to reach speed v. On Day 1, the total mass is 60 + 100 = 160 kg. On Day 2, adding the son brings the mass to 160 + 40 = 200 kg. Since speed is identical, the ratio of fuel is simply the ratio of their masses: 160 / 200 = 4 / 5.
The ratio of fuel used is 4:5 (or 0.8).
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.
According to the principle of levers (Load × Load Arm = Effort × Effort Arm), since the adult weighs twice as much as the child (2m × d = m × 2d), the adult must slide to exactly half the distance from the fulcrum compared to the child to keep the system perfectly balanced.
The gravitational attraction on the surface of the Moon (lunar surface) is about 1/6 th 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 ball will travel a height of 48 meters on the Moon.
Using the conservation of mechanical energy, the initial kinetic launch energy transforms completely into potential energy at the peak (½ mv² = mgh → h = v² / 2g). This proves height is inversely linked to gravity. Since the Moon's gravity pull is 6 times weaker, the ball floats up 6 times higher (8 × 6 = 48 m).
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 (ii) at 4 m. Does the block have negative acceleration in any portion of its motion?
- (i) Speed at 0 m: Since Kinetic Energy = 180 J, we set up 180 = ½ × 10 × v² → v² = 36 → 6 m/s.
- (ii) Speed at 4 m: The total work done is the area of the trapezoid under the graph: Area = ½ × (2 + 4) × 50 = 150 J. The total Kinetic Energy at 4 m becomes 180 + 150 = 330 J. Solving for speed: 330 = ½ × 10 × v² → v² = 66 → 8.12 m/s.
- Negative Acceleration? No. Even though the pushing force gets smaller between 3 m and 4 m, it is still a positive force pushing the block forward. The block continues to speed up the whole time, so acceleration never becomes negative.
The gravitational attraction on the surface of the Moon (lunar surface) is about 1/6 th 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 ball will travel a height of 48 meters on the Moon.
Using the conservation of mechanical energy, the initial kinetic launch energy transforms completely into potential energy at the peak (½ mv² = mgh → h = v² / 2g). This proves height is inversely linked to gravity. Since the Moon's gravity pull is 6 times weaker, the ball floats up 6 times higher (8 × 6 = 48 m).
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.
- (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?
- (i) Motion A-B: The flat line means the car moves at a perfectly constant speed of 35 m/s. This represents the brief reaction time split-second before the driver physically hits the brake pedal.
- (ii) KE at A: KE = ½ mv² = ½ × 1000 × 35² = 612,500 J.
- (iii) Work Done by Brakes: The brakes completely neutralize all forward motion down to zero, meaning the work done by braking friction is equal to −612,500 J.
- (iv) Transformation: The car's kinetic energy transforms completely into heat (thermal energy) and scraping sound energy inside the brake assemblies and tires.
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 s¹ and potential energy is 30 J. Calculate the velocity of the ball at P, Q and R.
Since the ball is stationary at O, its total mechanical energy budget is locked at 30 J (Kinetic Energy + Potential Energy = 0 + 30 = 30 J).
- At P (PE = 10 J): KE = 30 − 10 = 20 J. Solving for speed: 20 = ½ × 0.5 × v² → v² = 80 → 8.94 m/s.
- At Q (PE = 20 J): KE = 30 − 20 = 10 J. Solving for speed: 10 = ½ × 0.5 × v² → v² = 40 → 6.32 m/s.
- At R (PE = 40 J): The track graph at point R requires a potential energy of 40 J. Because our ball only has a total energy budget of 30 J, it physically cannot climb up to point R. It will stop and roll backward before ever reaching it. Velocity at R is impossible (mechanically unreachable).
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.
- (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.
- (i) Velocity before hitting: Using conservation of energy (½ mv² = mgh → v = √(2gh)), we compute v = √(2 × 10 × 10) = √(200) = 14.14 m/s.
- (ii) Sand Depression Depth: The initial potential energy at the top of the tree is 1.5 × 10 × 10 = 150 J. The work done by the sand stopping force must balance this energy (Force × distance = Energy): 3000 × d = 150 → d = 150 / 3000 = 0.05 meters (5 cm).