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  • A pulling force F = 50 N acting on a block of mass 5 Kg at an angle of 370 with horizontal as shown in the figure. the velocity of the block (in m/s)  after 5 sec will be (assume in

A pulling force F = 50 N acting on a block of mass 5 Kg at an angle of 370 with horizontal as shown in the figure. the velocity of the block (in m/s)  after 5 sec will be (assume initially block is at rest)

Option: 1

40


Option: 2

20


Option: 3

10


Option: 4

50


Answers (1)

best_answer

Solution :

\\ \text{Free body diagram of given block : }

\\ \text{Horizontal component of the force,}\left ( F_{x} \right ) =F\cos37^{\circ}=50\times \frac{4}{5}=40N \\ \text{Vertical component of the force,}\left ( F_{y} \right )=F\sin37^{\circ}=50\times \frac{3}{5}=30N \\ \text{Acceleration fo the block,(a)}=\frac{\text{component of force in the direction of acceleration}}{\text{mass}} \\ \\ a=\frac{40}{5}= 8 \ m/s^{2} \\ \\ \text{Now,from equation of motion} \\ v=u+at \\ v=0+8\times 5 =40 \ m/s

 

 

Posted by

himanshu.meshram

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