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Dafna1 [17]
3 years ago
10

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?​

Physics
1 answer:
lesantik [10]3 years ago
8 0

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let f be frictional force acting in the backward direction as shown. Let the coefficient of friction be \mu. Let N be the normal reaction force acting on the bag.

Given:

Mass of the bag is, m=8.10\textrm{ kg}

Force acting at \theta = 38° is F= 29.5\textrm{ N}

Acceleration due to gravity is, g=9.8\textrm{ }m/s^{2}

The force F can be resolved into its components as F_{x}=F \cos \theta and F_{y}=F \sin \theta

Therefore,

F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

f=F_{x}\\f=23.25\textrm{ N}

Now, frictional force is given as:

f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38

Therefore, the coefficient of friction is 0.38.

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