A 15.7 kg block is dragged over a horizontal surface by a constant force of 169 N acting at an angle of 31.4° above the horizontal. The block is displaced 11.6 m, and the coefficient of kinetic friction is 0.221.

Find the work done by the 169 N force. The acceleration of gravity is 9.8 m/s². Answer in units of J.

Find the magnitude of the work done by the force of friction. Answer in units of J.

Question

High School

Answer :

AngusHector AngusHector · Answer 1 · 2025-02-08T04:56:22-05:00

The work done by the 169 N force is 1678 J and the magnitude of the work done by the force of friction is 394.72 J.

To find the work done by the 169 N force, we need to calculate the horizontal component of the force and multiply it by the displacement.

The horizontal component of the force can be found using the formula:

F_horizontal = F * cos(theta)

F_horizontal = 169 N * cos(31.4°)

F_horizontal ≈ 144.6 N

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

Work = F_horizontal * displacement

Work = 144.6 N * 11.6 m

Work ≈ 1678 J

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

Work_friction = coefficient of friction * normal force * displacement

Normal force = mass * gravity

Normal force = 15.7 kg * 9.8 m/s²

Normal force ≈ 153.86 N

Work_friction = 0.221 * 153.86 N * 11.6 m

Work_friction ≈ 394.72 J

Therefore, the work done by the 169 N force is 1678 J and the magnitude of the work done by the force of friction is 394.72 J.

https://brainly.com/question/16987255

#SPJ1