Saved Bookmarks
| 1. |
Answer the following: (a) The casing of a rocket in flight burns up due to friction. At whose expense is the het energy required for burning obtained ? The rocked or the atmosphere ? (b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet's velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why? (c)An aritificial satellite orbiting the earth in very thin atmosphere loses its energy graduallydue to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth? (d) In figure, theman walks 2 m carrying a mass of 15 kg on his hands. In fig., he walks the same distance pulling the rope behind him. The rope goes over a pulley , and a mass of 15 kg hangs at its other end. In which case is the work done greater ? |
|
Answer» SOLUTION :(a) The total energy of a rocket in flight depends on its mass i.e., `P.E. +K.E. =mgh +(1)/(2) mv^(2).` When the casing burns up, its mass decreases. The total energy of the rocket decreases. Henc, heat energy required for burning is obtained from the rocket itselfand not from the atmosphere. (b) This is because gravitational force is a CONSERVATIVE force. Work done by the gravitational force of the sun over a closed path in every complete orbit of the comet is zero. (c) Whenthe artificial SATELLITE orbiting the earth comes closer and closer to earth, its potential energy decreases. As sum of potential energy and kinetic energy is constant, therefore, K.E. of satellite and hence its velocity goes on increasing. However, total energy of the statellite decreases a little on account of dissipation against atmospheric resistance. (d) In figure, force is applied on the mass, by the man is vertically UPWARD direction but distance is moved along the horizontal. `:. theta=90^(@). W=F s cos90^(@)=Zero . ` In figure, force is applied along the horizontal and the distance moved is also along the horizontal. Therefore,, `theta=0^(@).` `W=Fs cos theta =mg xx s cos 0^(@).` `w=15xx9.8 xx2xx1=294 jou l e. :. ` Work done in 2nd case is greater. |
|