14) calculate the work done in lifting 200kg of water of vertical height of 6 m?
15) prove KE = 1/2 mv^2?

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Solution

$D e a r s t u d e n t, i) w o r k d o n e = m a s s \times g \times h W = 200 \times 6 \times 10 = 12000 J i i) F r o m e q u a t i o n s o f m o t i o n, t h e r e l a t i o n c o n n e c t i n g t h e i n i t i a l v e l o c i t y (u) a n d f i n a l v e l o c i t y (v) o f a n o b j e c t m o v i n g w i t h a u n i f o r m a c c e l e r a t i o n a, a n d t h e d i s p l a c e m e n t, s f r o m N e w t o n ’ s s e c o n d l a w o f m o t i o n F = m a . w o r k d o n e b y t h e f o r c e, F i s w r i t t e n a s t h e f o l l o w i n g : w = m a \frac{v^{2} - u^{2}}{2 a} = \frac{1}{2} m (v^{2} - u^{2}) I f t h e o b j e c t i s s t a r t i n g f r o m i t s s t a t i o n a r y p o s i t i o n, t h a t i s, u = 0, t h e n w = \frac{1}{2} m v^{2} F r o m w o r k a n d e n e r g y t h e o r e m, w o r k d o n e i s e q u a l t o t h e c h a n g e i n t h e k i n e t i c e n e r g y o f t h e o b j e c t . K . E = \frac{1}{2} m v^{2} R e g a r d s$

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