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Two forces F1 and F2 are acting on a block of mass m=1.5 kg. The magnitude of
force F1 is 12N and it makes an angle= 37degree with the horizontal as shown in
figure-1. The block is sliding at a constant velocity over a frictionless floor. (a) Find the value of the normal force on the block. Find the magnitude of force F2 that is acting on the block. (c) Find the magnitude of force ~F2 if the block accelerates with a magnitude of a = 2:5 m=s2 along the direction of F2

Two forces F1 and F2 are acting on a block of mass m15 kg The magnitude of force F1 is 12N and it makes an angle 37degree with the horizontal as shown in figure class=

Respuesta :

Answer:

Explanation:

Component of force F₁ in right direction = F₁cos37

= 12 x cos37 = 9.58 N .

Component of force F₂ in  vertically upward direction = F₁sin37

= 12 x sin37 = 7.22 N .

a ) Let normal force be R

R + F₁sin37  = mg

R + 7.22 = 1.5 x 9.8 = 14.7

R = 7.48 N .

b )

Net force in horizontal direction = F₂ - F₁cos37

= F₂ - 9.58

This is equal to zero as body is moving with zero acceleration

F₂ - 9.58 = 0

F₂ = 9.58 N

c ) If body is moving with acceleration of 2.5 m /s² along the direction of F₂

F₂ - 9.58 = 1.5 x 2.5 = 3.75

F₂ = 13.33 N .

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