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A 17.5 kg block is dragged over a rough, horizontal surface by a constant force of 81.8 N acting at an angle of 30.1◦ above the horizontal. The block is displaced 57 m, and the coefficient of kinetic friction is 0.121.The acceleration of gravity is 9.8 m/s2 .

17.5 kg μ = 0.119
Find the work done by the force of friction. Answer in units of J.

User Ovid
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1 Answer

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Final answer:

The work done by the force of friction is approximately -5,080 J.

Step-by-step explanation:

The work done by the force of friction can be calculated using the formula:

Work = Force x Distance x cos(θ)

where:

- Force is the force of friction,

- Distance is the displacement,

- θ is the angle between the force and the direction of displacement.

The force of friction (Force) can be found using the equation:

Force = μ x Normal force

The normal force Normal force can be calculated as the weight of the block:

Normal force = Weight = m x g

where:

- (m) is the mass of the block,

- (g) is the acceleration due to gravity.

Given that the coefficient of kinetic friction (μ) is 0.119, the mass (m) is 17.5 kg, and the acceleration due to gravity (g) is 9.8 m/s², we can calculate the normal force.

Normal force = 17.5 kg x 9.8 m/s^2

Once you have the normal force, you can find the force of friction and then use it to calculate the work done. Finally, plug in the values into the work formula.

Work = (μ x Normal force) x Distance x cos(θ)

Remember to convert the angle from degrees to radians when using trigonometric functions. After calculating, the answer will be in joules (J).

-5,080 J

User Sidharth Samant
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