Hc Verma Work Power Energy Questions Pdf Download Class 9

An unruly demonstrator lifts a stone of mass 200 g from the ground and throws it at his opponent. At the time of projection, the stone is 150 cm above the ground and has a speed of 3 m/s. Calculate the work done by the demonstrator during the process. If it takes one second for the demonstrator to lift the stone and throw it, what horsepower does he use?

This page consists of detail step by step explanation of each and every question which is given in HC Verma physics, in the chapter Introduction of physics which is 1st chapter of HC Verma concept of physics in Volume -1 all questions are solved with proper explanation of all steps to enhance your learning.Read the theory of NCERT text book and solve questions given in the text book with the help of PhysicsWallah NCERT Solutions for class 11 Physics.

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The negative of the work done by the conservative internal forces on a system equals the change in:

\r\n1. total energy

\r\n2. kinetic energy

\r\n3. potential energy

\r\n4. none of these

The magnetic force on a charged particle is always perpendicular to its velocity. Therefore, the work done by the magnetic force on the charged particle is zero. Here, the kinetic energy and speed of the particle remain unaffected, while the velocity changes due to the change in direction of its motion.

(d) it moves in a circular path.

When the force on a particle is always perpendicular to its velocity, the work done by the force on the particle is zero, as the angle between the force and velocity is 90. So, kinetic energy of the particle will remain constant. The force acting perpendicular to the velocity of the particle provides centripetal acceleration that causes the particle to move in a circular path.

(a) the path taken by the suitcase

(b) the time taken by you in doing so

(d) your weight

Work done by us on the suitcase is equal to the change in potential energy of the suitcase.

i.e., W = mgh

Here, mg is the weight of the suitcase and h is height of the table.

Hence, work done by the conservative (gravitational) force does not depend on the path.

The total work done on a particle is equal to the change in its kinetic energy

(a) always

(b) only if the forces acting on it are conservative

(d) only if elastic force alone acts on it.

Consider two observers moving with respect to each other at a speed v along a straight line. They observe a bock of mass m moving a distance l on a rough surface. The following quantities will be same as observed by the two observers.

(a) kinetic energy of the block at time t

(b) work done by friction

(d) acceleration of the block.

A block of mass 2 kg kept at rest on an inclined plane of inclination 37 is pulled up the plane by applying a constant force of 20 N parallel to the incline. The force acts for one second. (a) Show that the work done by the applied force does not exceed 40 J. (b) Find the work done by the force of gravity in that one second if the work done by the applied force is 40 J. (c) Find the kinetic energy of the block at the instant the force ceases to act. Take g = 10 m/s2.

Given, m1=4 kg, m2=1 kg,v2=0.3 m/sv1=20.3=0.6 m/sv1=2v2 in this systemHeight descended by the 1 kg block, h=1 mDistance travelled by the 4 kg block,s=21=2 mInitially the system is at rest. So, u=0Applying work energy theoremwhich says that change in K.E.=Work done for the system12 m112+12 m222=-R s+m2gh

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