11th Physics chapter 5
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– 5.1 Introduction: Vectors are essential in physics for representing physical quantities like displacement, velocity, acceleration, and force. Understanding the scalar product of vectors is crucial in this chapter.
– 5.2 Notions of work and kinetic energy: Introduces the work-energy theorem, relating work done by forces to changes in kinetic energy. It defines work as the product of force and displacement, while kinetic energy is half the mass times the square of the speed【4:0†source】.
– 5.3 Work: Discusses the nature of work done by forces, distinguishing between positive and negative work and scenarios where work is calculated without knowing the exact force involved【4:1†source】.
– 5.4 Kinetic energy: Expands on the concept of kinetic energy and its relation to the work done on an object by forces. Kinetic energy is defined as half the mass times the square of the speed【4:2†source】.
– 5.5 Work done by a variable force: Examines scenarios where the work done by variable forces is calculated, including examples involving gravitational forces and resistive forces【4:3†source】.
– 5.6 The work-energy theorem for a variable force: Discusses the work-energy theorem and its application in situations where forces vary, linking changes in kinetic energy to the work done by net forces【4:2†source】.
– 5.7 The concept of potential energy: Introduces potential energy and its association with conservative forces, highlighting the undetermined nature of potential energy up to a constant【4:1†source】.
– 5.8 The conservation of mechanical energy: Explores the conservation of mechanical energy as a consequence of the work-energy theorem for conservative forces【4:1†source】.
– 5.9 The potential energy of a spring: Focuses on the potential energy stored in a spring system, with the equilibrium position defining the zero of potential energy for spring systems【4:1†source】.
– 5.10 Power: Defines power as the rate at which work is done and energy is transferred, introducing the concept of average and instantaneous power in relation to forces and velocities【4:4†source】.
– 5.11 Collisions: Applies the laws of momentum and energy conservation to study collisions between masses, emphasizing the relationship between masses, velocities, and angles in collisions【4:4†source】.
What is the work done by the resistive force in the entire journey if the speed of a raindrop on the ground is 10 m s-1?
The work done by the resistive force in the entire journey is zero since the raindrop has reached a uniform speed of 10 m s-1, indicating that it is no longer accelerating due to the resistive force of air.
What is the efficiency of the pump if it can pump up water to fill a tank of volume 30 m3 in 15 minutes, and how much electric power is consumed by the pump if the tank is 40 m above the ground?
The efficiency of the pump is 30%, and the electric power consumed by the pump is calculated as follows: Power = (Work done to fill the tank) / (Time taken) = (Potential energy of water in the tank) / (Time taken) = (mgh) / t = (30000 kg * 9.8 m/s^2 * 40 m) / (900 s) = 117.33W.
If two identical ball bearings in contact with each other are hit head-on by another ball bearing of the same mass moving initially with a speed V in an elastic collision, what is a possible result after the collision?
One possible result after an elastic collision is the transfer of kinetic energy from the moving ball bearing to the two stationary ball bearings, causing the moving ball bearing to stop and the other two ball bearings to move with a shared kinetic energy and velocity.
When a trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly on a frictionless track at 27 km/h and sand starts leaking out at a rate of 0.05 kg/s, what is the speed of the trolley after the entire sandbag is empty?
The speed of the trolley after the entire sandbag is empty will remain unaffected at 27 km/h since the leaking sand does not affect the overall momentum of the system on the frictionless track.
In a linear motion where a body of mass 0.5 kg travels with velocity v = a x^3/2 and a = 5 m^(-1/2) s^(-1), what is the work done by the net force during its displacement from x = 0 to x = 2 m?
The work done by the net force during the displacement is calculated by integrating the force over the displacement range, given by the formula: Work = Integral [F(x) * dx] for x = 0 to x = 2. Substituting the given values and integrating the force equation gives the work done.
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