SATHEE: Work Energy And Power Question 247

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 $μ_{k} = 0.5$ . 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. $m g (8) - μ m g c o s 53^{\circ} (10) - u m g (10) - μ m g c o s$

$37^{\circ} (x)$

$- m g s i n 37^{\circ} (x) = 0 - \frac{1}{2} (1) {(10)}^{2}$ On solving we get $x = 1 m$

Work Energy And Power Question 247

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Work Energy And Power Question 247

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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