A 10.0kg box being pushed at constant velocity against a 3.0N[left] force offriction.a. Draw free body diagram (2 marks)b. Find the coefficient of friction. State if it is static or kinetic. (2 marks)
Question
A 10.0kg box being pushed at constant velocity against a 3.0N[left] force offriction.a. Draw free body diagram (2 marks)b. Find the coefficient of friction. State if it is static or kinetic. (2 marks)
Solution
a. Free Body Diagram:
The free body diagram would include the following forces:
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The force of gravity (Fg) acting downwards. This can be calculated as mass (m) times gravity (g). For a 10.0 kg box, this would be 10.0 kg * 9.8 m/s^2 = 98 N [downwards].
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The normal force (Fn) acting upwards. In this case, since the box is not accelerating vertically, the normal force equals the force of gravity, so Fn = 98 N [upwards].
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The force of friction (Ff) acting to the left. This is given as 3.0 N [left].
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The applied force (Fa) acting to the right. Since the box is moving at a constant velocity, the applied force equals the force of friction, so Fa = 3.0 N [right].
b. Coefficient of Friction:
The coefficient of friction (μ) can be found using the formula Ff = μFn. We can rearrange this to find μ = Ff / Fn.
Substituting the given values, we get μ = 3.0 N / 98 N = 0.0306.
Since the box is moving at a constant velocity, this is the coefficient of kinetic friction.
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