Asked by Morty

A 2.0Kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in .50s. What net force acts on the otter along the incline?

Answers

Answered by drwls
The force (F) equals the mass (M= 2.0 kg) times the acceleration.

If it travels 0.85 m in t = 0.50 s, the acceleration a is given by solving
0.85 = (1/2) a t^2 = = (a/2)*0.25
a = 6.8 m/s^2

Solve for F
Answered by tyreek
13.6 n
Answered by Hala
givens:M=2.0kg,vi=0m/s,displacement=85cm(0.85m),t=0.50s.
formula=average acceleration=average velocity/total time
>>average velocity=displacement/time
>>average velocity=0.85/0.5=1.7m/s
>>average acceleration=1.7/0.5=3.4m/s^2
force=mass*acceleration
force=2.0*3.4=6.8N
Answered by Cyrus B. Vonyon
6.8n

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