Physics
posted by Morty on .
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?

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 
13.6 n

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