Work Energy And Power Question 247
Question: A 1 kg particle at a height of 8m has a speed of 1 Om/s down a fixed incline making an angle $ 53{}^\circ $ with horizontal as shown in figure. It slides on a horizontal section of length 10 m at ground level and then up a fixed incline making an angle $ 37{}^\circ $ with horizontal. All surfaces have $ {\mu _{k}} = 0.5 $ . How far (in meters) from point 0 (bottom of right inclined plane), along the incline making an angle $ 37{}^\circ $ with horizontal, does the particle first come to rest?
Options:
A) 1m
B) 0.5m
C) 2m
D) 5m
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Let x be the distance travelled by the particle along the $ 37{}^\circ $ incline, after which it comes to rest. From work-energy theorem. Work done by gravity on block on incline $ 53{}^\circ + $ work done by friction on block on incline $ 53{}^\circ = $ work done by friction on ground level + work done by gravity on block on incline $ 37{}^\circ + $ work done by friction on block on incline $ 37{}^\circ = $ change in K.E. $ mg( 8 )-\mu mg cos 53{}^\circ ( 10 ) - umg( 10 )-\mu mg cos $
$ 37{}^\circ ( x ) $
$ -mg sin 37{}^\circ ( x ) = 0-\frac{1}{2}( 1 ) {{( 10 )}^{2}} $ On solving we get $ x=1m $
BETA
