Question

what is the external net force exerted on a 3.5 kg papaya, which is being pushed across a table and has an acceleration of 2.2 m /s² to the left​

Answer 1

Answer:

Given -

• Mass (m) = 3.5 kg
• Acceleration (a) = 2.2 m/s²

To find -

• External net force exerted on papaya.

Solution -

From the second law of motion i.e., F = MA

→ F = 3.5 × 2.2

→ 7.7 kg ms-²

Therefore, the external net force exerted on papaya is 7.7 kg m/s².