First slide

Work

Question

A block of mass 10kg is moving in x direction with a constant speed of 10m/s. It is subjected to a retarding force F=-0.1xJ/m during its travel from x=20m to x=30m. Its final kinetic energy will be

Moderate

A

499J
B

450 J
C

275 J
D

250 J

Solution

$w o r k d o n e = c h a n g e i n K E F . S = K E_{f i n a l} - K E_{i n i t i a l} K E_{f i n a l} = F . S + K E_{i n i t i a l} (h e r e r e t a r d i n g f o r c e i s n e g a t i v e) s u b s t i t u t e g i v e n v a l u e s K E_{f i n a l} = (- 0.1) (30 - 20) + \frac{1}{2} (10) 10^{2} K E_{f i n a l} = 499 j o u l e - - - - - - - - - - - - - - - - - - - - - s e c o n d m e t h o d : i n i t i a l v e l o c i t y = u = 10 m / s a c c e l e r t i o n a = - \frac{F}{m} = - \frac{0.1}{10} = - 0.01 m / s^{2} d i s p l a c e m e n t s = 30 - 20 = 10 m f i n a l v e l o c i t y v = ? e q u a t i o n o f m o t i o n i s v^{2} - u^{2} = 2 a s s u b s t i t u t e g i v e n v a l u e s v^{2} = 10^{2} - 2 (0.01) 10 \Rightarrow v^{2} = 99.8 - - - (1) K E_{f i n a l} = \frac{1}{2} m v^{2} s u b s t i t u t e e q n (1) a n d g i v e n m a s s m = 10 k g K E_{f i n a l} = \frac{1}{2} (10) (99.8) = 499 J$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App