Question #4874

A 2.3 kg otter starts from rest at the top of a

muddy incline 90.5 cm long and slides down

to the bottom in 0.50 s.

What net external force acts on the otter

along the incline?

Answer in units of N

muddy incline 90.5 cm long and slides down

to the bottom in 0.50 s.

What net external force acts on the otter

along the incline?

Answer in units of N

Expert's answer

Let's use the second Newton's law:

F = ma.

All we need is to calculate acceleration a. Why can do it in a such way:

Va = 0.905/0.50 = 1,81 m/s - average speed of an ottet.

We know that V(0) = 0. Also we& know that V(0.5/2) = V(0.25) = Va.

So, a = (V(0.25)-V(0))/0.25 = (Va-0)*4 = 4*Va = 4*1.81 = 7,24 m/s^2.

At last we obtain F = ma = 2.3*7.24 = 16,652 N.

F = ma.

All we need is to calculate acceleration a. Why can do it in a such way:

Va = 0.905/0.50 = 1,81 m/s - average speed of an ottet.

We know that V(0) = 0. Also we& know that V(0.5/2) = V(0.25) = Va.

So, a = (V(0.25)-V(0))/0.25 = (Va-0)*4 = 4*Va = 4*1.81 = 7,24 m/s^2.

At last we obtain F = ma = 2.3*7.24 = 16,652 N.

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