A 1.7 kg otter starts from rest at the top of a
muddy incline 89.2 cm long and slides down
to the bottom in 0.60 s.
What net external force acts on the otter
along the incline?
Answer in units of N
external force? gravity, friction.
vavg= .892m/.60s
Vf= 2*vavg
a= Vf/.6
forceactingnet=mass*a
To find the net external force acting on the otter along the incline, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Step 1: Find the acceleration of the otter along the incline.
To find the acceleration, we can use the following kinematic equation:
vf^2 = vi^2 + 2ad
Where vf is the final velocity, vi is the initial velocity (which is 0 m/s as the otter starts from rest), a is the acceleration, and d is the distance the otter slides down the incline.
Rearranging the equation, we have:
a = (vf^2 - vi^2) / (2d)
vf = (2d) / t [Using the formula vf = d/t, as the otter covers the distance in the given time]
Plugging in the given values:
vf = (2 * 0.892 m) / 0.60 s ≈ 2.98 m/s
a = [(2.98 m/s)^2 - (0 m/s)^2] / (2 * 0.892 m)
a ≈ 5.98 m/s^2
Step 2: Calculate the net external force.
Using Newton's second law,
Fnet = m * a
Plugging in the given mass and calculated acceleration:
Fnet = (1.7 kg) * (5.98 m/s^2)
Fnet ≈ 10.17 N
Therefore, the net external force acting on the otter along the incline is approximately 10.17 N.