No as m ass and velocity cannot ne negative
Q.2. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
A freely falling object just before hitting the ground has maximum kinetic energy. After falling, it rolls on the rough ground and finally comes to rest. The kinetic energy of the object is used up in doing work against friction; which finally appears as heat energy.
Q.3. Find the energy in kWh consumed in 10 hours by four devices of power 500 W each.
Energy consumed = Power x time taken = 2000W x 10 h = 20000 Wh = 20 kWh.
Q.4. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
The work required to be done to stop a car = ( ½ mu2) - ( ½ mv2)= ½ m(u2-v2)= 1/2x1500(602-0)=2.08J
Q.5 What is the work done by the force of gravity on a satellite moving round the Earth? Justify your answer.
The work done by the force of gravity on a satellite moving around the Earth is zero.
When a satellite moves around the Earth in a circular path, then the force of gravity acts on it directed towards the centre. The motion of the satellite is in the horizontal plane. Therefore, the force of gravity of Earth on the satellite and the direction of motion of satellite are perpendicular to each other. Therefore, net work done = Fs cos 90 = 0.
Q.6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
During the free fall of the object, there is continuous decrease in potential energy. This decrease in potential energy appears as an equal amount of increase in kinetic energy. Thus, the sum of the potential energy and kinetic energy of the object would be the same at all points. That is, potential energy + kinetic energy = constant.
According to the law of conservation of energy, the total energy of system remains unchanged. Thus, the given statement does not violate the law of conservation of energy.
Define 1 J of work.
Work done = Force x Displacement
If force, F = 1 N and displacement, s = 1, m then the work done by the force will be 1 Nm or1 J. Thus, 1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.
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