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  • A block of mass m = 5 Kg is moving up on the inclined plane of angle  with horizontal with constant ve

A block of mass m = 5 Kg is moving up on the inclined plane of angle \theta = 30 \degree with horizontal with constant velocity V by the help of force F acting along the plane and upwards. What is the work (in Joule) done by force F to move the object up the inclined by distance 's'  [ given \mu_k = 1/ \sqrt 3]  b/w surface and block  and  s = 2 m 

g = 10 m/s ^2

Option: 1

50


Option: 2

75


Option: 3

80


Option: 4

100


Answers (1)

As we have learned

Positive Work -

0\leq \Theta < \frac{\pi }{2}

- wherein

\Theta is the angle between force vector and displacement vector

 

 

W = FS \cos \phi

\phi = 0 \degree

be balancing force along inclined plane 

F = mg \sin \theta + f_k

F = mg \sin \theta + \mu _k mg \cos \theta

w= mg ( \sin \theta + \mu \cos \theta ) \times s \times \cos 0

= 5 \times 10 ( 1/2 + \frac{1}{\sqrt 3} \frac{\sqrt 3 }{2}) \times 2 \times 1 = 100 J

 

 

 

 

 

 

 

 

 

 

 

 

 

Posted by

Kshitij

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