A Block Is Moved Up A 600 Incline As Shown In The Figure

A Block Is Being Moved Up A 600 Incline As Shown In The Figure Above

A block is being moved up a 60.0° incline. The block has a mass of 16.0 kg. The block is being pushed by an applied force Fapp = 2510 N which is directed perfectly horizontally. The normal force on the block is FN = 2179 N. The block is subject to kinetic friction characterized by a coefficient of friction equal to 0.270. The block moves a distance d = 1.10 m along the ramp as sketched in the figure. For each force listed below, find the work done during the motion of the block. Be sure to include the appropriate plus or minus sign. For each force listed below, find the work done during the motion of the block. Be sure to include the appropriate plus or minus sign.

Applied force

Force of gravity

Normal force

Kinetic friction force

Find the net work of the block

Paper For Above instruction

The movement of a mass along an inclined plane involves understanding the work done by various forces acting on the object. In this scenario, a 16.0 kg block is being pushed up a 60° incline over a distance of 1.10 meters, with specific forces applied and significant resistive effects such as friction and gravity. Calculating the work done by each force requires decomposing these forces into components parallel or perpendicular to the incline and considering the direction of each force relative to the displacement.

Parameters and Given Data

  • Mass of the block, m = 16.0 kg
  • Incline angle, θ = 60.0°
  • Applied force, Fapp = 2510 N (horizontal)
  • Normal force, FN = 2179 N
  • Coefficient of kinetic friction, μk = 0.270
  • Displacement along the incline, d = 1.10 m

Additional calculations involve resolving the applied force into components relative to the incline, the gravitational component along the incline, and the frictional force, which opposes motion. The work done by each force = force component × displacement × cosine of the angle between the force and the displacement direction. Since the forces are aligned or oppose each other along the displacement, the cosine of the angle will be accordingly 1 or -1 for positive or negative work.

Calculation of Individual Works

1. Work Done by the Applied Force (Fapp)

The applied force is horizontal, while the motion is along the incline. The component of the applied force along the incline is Fapp_x = Fapp * cos(α), where α is the angle between the applied force direction and the incline.

Considering the geometry:

- The applied force is horizontal.

- The incline makes an angle of 60°, and the applied force acts horizontally.

- The component of Fapp along the incline is: Fapp_parallel = Fapp × cos(θ) = 2510 N × cos(60°) = 2510 × 0.5 = 1255 N.

The work done by this force as the block moves up is:

W_apply = Fapp_parallel × d = 1255 N × 1.10 m = 1380.5 J.

Since the force component acts in the same direction as the displacement, the work done is positive:

W_apply = +1380.5 J.

2. Work Done by Gravity (Force of Gravity)

Gravity acts vertically downward with magnitude:

F_gravity = m × g = 16.0 kg × 9.81 m/s² ≈ 157.0 N.

The component of gravity along the incline:

F_gravity_parallel = F_gravity × sin(θ) = 157.0 N × sin(60°) ≈ 157.0 × 0.866 = 136.0 N.

This component acts downward, opposite to the movement direction, so:

Work done by gravity:

W_gravity = - F_gravity_parallel × d = -136.0 N × 1.10 m ≈ -149.6 J.

W_gravity ≈ -149.6 J.

3. Work Done by Normal Force (FN)

Normal force acts perpendicular to the incline surface; since the displacement is along the incline and the normal force is perpendicular to that movement, the work done by the normal force is zero:

W_normal = 0 J.

4. Work Done by Kinetic Friction (F_friction)

The frictional force magnitude:

F_friction = μk × FN = 0.270 × 2179 N ≈ 588.33 N.

This force opposes the direction of motion, so the work done:

W_friction = - F_friction × d = -588.33 N × 1.10 m ≈ -647.2 J.

W_friction ≈ -647.2 J.

5. Net Work Done on the Block

The net work is the sum of all individual works:

W_net = W_apply + W_gravity + W_normal + W_friction

W_net ≈ 1380.5 J - 149.6 J + 0 J - 647.2 J

W_net ≈ (1380.5 - 149.6 - 647.2) J ≈ 583.7 J.

The net work done on the block is approximately 583.7 J.

Conclusion

The calculations show that despite resistive forces like gravity and friction opposing the movement, the applied force provides enough work to produce a net positive work of approximately 584 Joules, accelerating the block up the incline. These calculations are essential in understanding energy transfer and the dynamics of objects moving along inclined planes under various forces.

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