Question

A net force of 6.6N east acts on a 9.0kg object. If this object accelerates uniformly from rest to a velocity of 3.0m/s east. How far did the object travel and how long did it take?​

Answer 1

Given :-

☄ Net Force, f = 6.6 N

☄ Mass of the object, m = 9 kg

☄ Initial velocity, u = 0 m/s

☄ Final velocity, v = 3 m/s

To find :-

How far did the object travel and how long did it take

Solution :-

using Newton's law of motion .i.e.,

➠ F = ma

here, f = force, m = mass and a = accerlation

➠ 6.6 = (9)(a)

➠ a = 6.6/9

➠ a = 0.73 m/s²

now,

using first equation of motion .i.e.,

➠ v = u + at

➠ 3 = 0 + 0.73t

➠ t = 3/0.73

➠ t = 4.1 s

thus, the time taken by the object is 4.1 s

Now, using 2nd equation of motion .i.e.,

➠ s = ut + 1/2 at²

➠ s = (0)(4.1) + 1/2(0.73)(4.1)²

➠ s = 0.73 × 8

➠ s = 5.8 m.

thus, the distance traveled by the object is 5.8 m

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