A 3.30 kg box is moving to the right with speed 10.0 m/s on a horizontal, fricti

Question

A 3.30 kg box is moving to the right with speed 10.0 m/s on a horizontal, frictionless surface. At t = 0 a horizontal force is applied to the box. The force is directed to the left and has magnitude F(t) = (6.00 N/s^2 )t^2 What distance does the box move from its position at t = 0 before its speed is reduced to zero? Express your answer with the appropriate units. x=16.3 m If the force continues to be applied, what is the velocity of the box at 3.00s ? Express your answer with the appropriate units. v_x = Value Units

Explanation / Answer

A)

We have,

F = ma

S = ut +(1/2)at2

v = u + at

Where,

F - Force = - 6 N

m - mass of the body = 3.3 kg

a - acceleration = ?

S - displacement = ?

u - initial velocity = 10 m/s

v - final velocity = 0 m/s

t - time taken = ?

Now,

F = ma

a =F/m = -6/3.3 = -1.82 m/s2(saying the force is being applied in the negative direction.)

Also, we know that

v = u + at

By using this we can find how long it takes to stop.

0 = 10 -1.82t

1.82t = 10

t = 5.494 s

Now, we can find the distance travelled before it stops.

S = ut +(1/2)at2 = 10 x 5.494 + (1/2) x (-1.82) x 5.4942 = 27.47 m

B) Now, when t = 3 s

v = u + at = 10 + (-1.82 x 3) = 4.54 m/s

Also, after the box stops also if we continue the force, the box will move in the opposite direction with a velocity v.

when the box stops velocity becomes zero. so u = 0 and t = 3s (counted from the starting of opposite movement)

v = u + at = 0 + (-1.82 x 3) = - 5.46 m/s

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