A 5.5 kg box is pushed across the lunch table. the net force applied to box is 9.7 N. what is the acceleration of the box?

Question

Physics

Aarav Boyd

1 September, 05:07

A 5.5 kg box is pushed across the lunch table. the net force applied to box is 9.7 N. what is the acceleration of the box?

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Answers (1)

Know the Answer?

Aidyn Zuniga · Answer 1

a = 1.76m/s^2

Explanation:

Force = mass x acceleration

F = m x a

N = kg x m/s^2

From the question

m = 5.5kg

F = 9.7N

Therefore,

Using the above formula

9.7 = 5.5 x a

9.7 = 5.5a

Divide both sides by 5.5

9.7/5.5 = 5.5a/5.5

1.76 = a

a = 1.76m/s^2