A crate of mass 30.0 kg is pulled by a force of 300 N up an inclined plane which makes an angle of 30º with the horizon. The coefficient of kinetic friction between the plane and the crate is μk = 0.225. If the crate starts from rest, calculate its speed after it has been pulled 15.0 m. Draw the free body diagram.

Introduction

In this problem, we are asked to calculate the final speed of a crate being pulled up an inclined plane with friction. To solve this, we apply Newton’s laws of motion and the work-energy theorem. We’ll also include a descriptive explanation of the free body diagram (FBD).

Given Data

  • Mass of crate, m = 30.0 kg
  • Force applied, F = 300 N
  • Distance moved, s = 15.0 m
  • Angle of incline, θ = 30º
  • Coefficient of kinetic friction, μk = 0.225
  • Initial speed, u = 0 m/s (starts from rest)
  • g = 9.8 m/s²

Step 1: Understanding the Free Body Diagram (FBD)

The forces acting on the crate are:

  • Gravitational force (mg) acting vertically downward
  • Normal force (N) perpendicular to the plane
  • Frictional force (f) acting opposite to the direction of motion (down the incline)
  • Applied force (F) up the incline

Frictional force: f = μk × N = μk × mg cos(θ)

Component of weight down the incline: mg sin(θ)

Step 2: Net Work Done

Using the Work-Energy Theorem:

Net Work Done = Change in Kinetic Energy

Let’s calculate the net force along the incline:

F_net = Applied force – frictional force – component of weight

F_net = F – (μk × mg × cosθ) – (mg × sinθ)

F_net = 300 – (0.225 × 30 × cos30º × 9.8) – (30 × sin30º × 9.8)

cos30º ≈ 0.866, sin30º = 0.5

F_net = 300 – (0.225 × 30 × 0.866 × 9.8) – (30 × 0.5 × 9.8)

F_net ≈ 300 – (0.225 × 30 × 8.487) – (147)

≈ 300 – (57.3) – (147) = 95.7 N

Step 3: Work-Energy Theorem

Net Work = F_net × s = 95.7 × 15 = 1435.5 J

Change in K.E. = ½ × m × v² – 0

So, 1435.5 = 0.5 × 30 × v²

v² = (1435.5 × 2)/30 = 95.7

v = √95.7 ≈ 9.78 m/s

Answer

The final speed of the crate after moving 15.0 m is approximately 9.78 m/s.

Free Body Diagram Description

  • Weight (mg): Acts vertically downwards.
  • Normal force (N): Acts perpendicular to the incline, balancing the perpendicular component of the weight.
  • Frictional force: Acts parallel to the incline, in the opposite direction of motion.
  • Applied force (F): Acts up the incline.

Conclusion

This problem demonstrates the use of forces, components, and energy concepts in mechanics. By applying the work-energy theorem, we find that the crate reaches a speed of about 9.78 m/s after moving 15 meters up the incline, considering both gravity and frictional resistance.

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