Question

What is the external force oexerted on3.5kg papaya which is being pushed across a table and has accelaration of 2.2m/s to the left

Answer 1

Answer:

Mass (m) = 3.5 kg

Acceleration (a) = 2.2 m/s²

From the second law of motion,

F = MA  

→ F = 3.5 × 2.2  

→ 7.7 kg ms-²

Hence the force applied on the papaya will be 7.7 kg ms-²

Explanation:

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