A 3.63.kgkg chihuahua charges at a speed of 3.3m/s3.3m/s. What is the magnitude of the average force needed to bring the chihuahua to a stop in 0.50s0.50s

23.96 N

Explanation:

From the question given above, the following data were obtained:

Mass of Chihuahua (m) = 3.63 kg

Velocity (v) = 3.3m/s

Time (t) = 0.50 s

Force (F) =?

Next, we shall determine the acceleration of the Chihuahua. This can be obtained as follow:

Velocity (v) = 3.3m/s

Time (t) = 0.50 s

Acceleration (a) =?

a = v/t

a = 3.3/0.5

a = 6.6 m/s²

Thus, the acceleration of the Chihuahua is 6.6 m/s².

Finally, we shall determine the force need to stop the Chihuahua as shown below:

Mass of Chihuahua (m) = 3.63 kg

Acceleration (a) = 6.6 m/s².

Force (F) =?

F = ma

F = 3.63 × 6.6

F = 23.96 N

Therefore, a force of 23.96 N is needed to stop the Chihuahua.

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