Respuesta :
Answer:
The net work done by friction on the body of the snake is 333.35J
Explanation:
Work done is given by
W = F × s
Where W is the Work done
F is the force
and s is the distance covered
Since we are to determine the work done by friction, then we will determine the frictional force. The frictional force is given by
f = μN = μw
Where μ is the coefficient of friction
N is the normal reaction
and w is the weight
But, F = f
∴ W = μws
From the question
μ = 0.25
w = 54.0 N
Now, we will determine s
From the question,
We are to determine the work done by friction on the body of a snake slithering in a complete circle of 3.93 m radius.
The distance s here is given by the circumference of the circle. Circumference of a circle is given by 2πr
∴ s = 2πr
s = 2 × π × 3.93
s = 7.86π m
Hence,
W = 0.25 × 54.0 × 7.86π
W = 333.35 J
Hence, the net work done by friction on the body of the snake is 333.35J.
The net work done by friction on the body of the snake is :
-333.35J
Friction
Formulas:
Work done (W) = F × s
F = force
s = distance covered
f = μN = μw
μ = coefficient of friction
N = normal reaction
w = weight
Solution:
F = f
Weight is :
W = μws
μ = 0.25
w = 54.0 N
Distance covered:
s = 2πr
s = 2 × π × 3.93
s = 7.86π m
Therefore,
W = 0.25 × 54.0 × 7.86π
W = 333.35 J
The net work done by friction on the body of the snake is 333.35J.
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