Question

A 16.3 kg block is dragged over a rough, horizontal surface by a constant force of 184 N acting at an angle of angle 34.9 above the horizontal. The block is displaced 44 m and the coefficient of kinetic friction is 0.131. 16.3 kg µ = 0.131 184 N 34.9 . Find the work done by the 184 N force. The acceleration of gravity is 9.8 m/s²

Answer 1

The work done by the 184 N force acting at an angle of 34.9 degrees above the horizontal over a displacement of 44 m is approximately 6711.76 joules.

Step-by-step explanation:

To find the work done by the 184 N force, we first calculate the horizontal component of the force since work is the product of the force in the direction of displacement and the magnitude of the displacement. The horizontal force (Fcos) is given by F * cos(θ), where F is the applied force and θ is the angle of the force above the horizontal. Here it would be 184 N * cos(34.9°).

Next, work done by the force (W) is the product of this horizontal force component and the displacement (d), so:

W = Fcos * d

Substituting the values:

W = (184 N * cos(34.9)) * 44 m

Remember to use cosine to convert to degrees in your calculator:

W ≈ (184 N * 0.829) * 44 m ≈ (152.54 N) * 44 m ≈ 6711.76 J

Therefore, the work done by the 184 N force over the 44 m displacement is approximately 6711.76 joules.