
#1
Nov409, 04:45 PM

P: 3

1. The problem statement, all variables and given/known data[/B]
A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides down to the bottom in 0.50 s. What net force acts on the otter along the incline? mass = 2.0 kg distance = 85 cm time = .50 s net force = ? 2. Relevant equations Fnet=ma I also used. V=x/t a=v/t 3. The attempt at a solution v=x/t 85/.5=170 a=v/t 170/.5=340 cm/s^2 340 cm/s^2 = 3.4 m/s^2 Fnet=ma (2.0 kg)(3.4 m/s^2)= 6.8 N The book says the right answer is 14 N though. 



#2
Nov409, 06:01 PM

P: 122





#3
Nov409, 08:58 PM

P: 2

Hi elkosp16, you need to use the formula :
Δ X = Vi*T + (1/2)*a*T^2 You already know displacement (ΔX), 85 cm (remember to convert it to meters !) Initial velocity (Vi) is 0, because it starts at rest. Time = .50 seconds, Plug everything in and solve for a, once you get acceleration, use newtons second law. ∑ F=m*a plug in mass & acceleration, the answer is net force. 


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