physics help plz?

Question

A 2.0kg 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?

Answer 1

13.6 n

Answer 2

Oops! Forgot a factor of .5. Assuming constant acceleration:

xf = x0 + v0 t + 0.5 a t^2

85 cm = 0 cm + 0 cm/s * 0.50 s + 0.5 * a * (0.50 s)^2

a = 680 cm/s^2 = 6.8 m/s^2

F = m * a = 2.0 kg * 6.80 m/s^2 = 13.6 N

The incline is "muddy", so it's incorrect to assume that the angle of the incline is any particular value since you don't know how much friction there is.

Answer 3

let assume that the angle of the plane is θ

if the otter want to climb with zero acceleration (constant velocity) he must act with o force

F = G sin θ

because the gravity will act with a force to stop the otter climbing which is equal to G sin θ
G = m g - weight
g = 9.81 m / s^2

F = m g sin θ

we must find θ ?

when he is in the top and sides down he has and accelerate motion with and acceleration

a = g sin θ

we also know that in accelerate motion

x = 0.5 a t^2

x = 0.5 g sin θ t^2

sin θ = 2 x / (g t^2)

x = 0.85 m
t = 0.5 s

sin θ = 0.693

F = m g sin θ

F = 2 * 9.81* 0.693 N

F = 13.6 N

Answer 4

F = ma = 2.0 * 9.8 = 19N (to 2 sig figs)