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A 2.00-kg block of ice is moving on a frictionless horizontal surface. At t = 0 the block is moving to the right with a velocity of magnitude 3.00 m/s. (a) Calculate the velocity of the block (magnitude and direction) after a force of 5.00 N directed to the right has been applied for 4.00 s; (b) if instead a force of 7.00 N directed to the left is applied from t=0 to t = 4.00 s, what is the final velocity of the block?

Respuesta :

Answer:

Explanation:

a )

We shall apply the concept of impulse .

Impulse = force x time = change in momentum

= 5 x 4 = 2 ( V - 3 )  , where V is final velocity of the object

20 = 2V - 6

V = 13 m /s

b )

Impulse applied = - 7 x 4 = - 28 kg m/s ( negative as direction of force is opposite motion )

If v be the final velocity

2 x 3 - 28 = 2 v  ( initial momentum - change in momentum = final momentum )

- 22 = 2v

v = - 11 m /s

object will move with 11 m /s in opposite direction .

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