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rosijanka [135]
4 years ago
8

3.6-kg chihuahua charges at a speed of 3.3 m/s what is the magnitude of the average force needed to bring the chihuahua to a sto

p in 0.50s?
Physics
1 answer:
patriot [66]4 years ago
8 0
Physics is my weakest subject, but it's worth a shot here. Hopefully the answer is right.

velocity(final) = velocity(initial) + acceleration * time
vf = vi + a * t
We want to bring the chihuahua to a stop in half a second. So, velocity(final) = 0 m/s. Its velocity(initial) = 3.3 m/s.

Solving for a, the acceleration.
0 m/s = 3.3 m/s + (a) * 0.5 s
a = -6.6 m/s^2

Force = mass * acceleration
F = m * a
We know the mass of the chihuahua and we just found the necessary acceleration.
Plug those in.
F = (3.6 kg) * (-6.6 m/s^2)

F = -23.76 N
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