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A ball is thrown vertically upward with a velocity of 19.6 metre per second and momentarily come to rest in how much time?.... |
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Answer» tion:Given that,Initial velocity u=19.6m/sFinal velocity v=0Time t=6secThe acceleration isWe know that, v=u+at 0=19.6+a×6 a=−3.26m/s 2 Now, the height isFrom EQUATION of motion s=ut+ 21 at 2 s=19.6×6− 21 ×3.26×36 s=58.9 s=59mHence, the height of the tower is 59 MTHE work–energy theorem: - The work done by all forces on an object is equal to the change in kinetic energy of the object. W=ΔkFrom newton second LAW,F=ma....(I)We know that,a= dtdv So, put the value of a in equation (I)F=m dtdv If multiple both SIDE by vF.v=mv dtdv .....(II)We know that,v= dtdx Now, put the value of v in equation (II)F dtdx =mv dtdv Now, on integrating Fdx=mvdv F 0∫x dx=m v 1 ∫v 2 vdv Fx= 21 m(v 22 −v 12 ) Fx= 21 mv 22 − 21 mv 12 We know that, W=Fx 21 mv 2 =kk = kinetic energyNow, W=k 2 −k 1 W=ΔkThe work –energy theorem is proved. |
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