To tackle this problem, we need to break it down into several parts: calculating the work done by the applied force, determining the mechanical energy lost during the climb, and finding the normal force acting on the slug block. Let's go through each step methodically.
Calculating Work Done by the Applied Force
The work done by a force is calculated using the formula:
Work = Force × Distance × cos(θ)
In this case, the force applied is 200 lbs, and the distance moved along the incline is 10 feet. Since the force is horizontal and the incline is at an angle of 60 degrees, we need to consider the angle between the force and the direction of movement. Here, θ is 0 degrees because the force is applied horizontally while the block moves up the incline.
Thus, the work done can be calculated as:
Work = 200 lbs × 10 ft × cos(0°) = 2000 ft-lbs
Determining Mechanical Energy Lost
Next, we need to find out how much mechanical energy was lost during the climb. This can be calculated by considering the change in kinetic energy and the potential energy gained by the block.
The change in kinetic energy (ΔKE) is given by:
ΔKE = 0.5 × m × (v_f² - v_i²)
Where:
- m = mass of the block = 3 kg (which we need to convert to slugs for consistency with the force unit)
- v_f = final velocity = 2 ft/sec
- v_i = initial velocity = 0 ft/sec
To convert mass from kilograms to slugs, we use the conversion factor: 1 kg = 0.0685 slugs. Therefore, 3 kg = 0.2055 slugs.
Now, substituting the values:
ΔKE = 0.5 × 0.2055 slugs × (2 ft/sec)² = 0.5 × 0.2055 × 4 = 0.411 ft-lbs
Next, we calculate the potential energy gained (PE) as the block climbs the incline:
PE = m × g × h
Where:
- g = acceleration due to gravity = 32.2 ft/sec²
- h = height gained = d × sin(θ) = 10 ft × sin(60°) = 10 ft × (√3/2) ≈ 8.66 ft
Now substituting the values:
PE = 0.2055 slugs × 32.2 ft/sec² × 8.66 ft ≈ 57.73 ft-lbs
The total mechanical energy input is the work done by the force, which is 2000 ft-lbs. The total mechanical energy output is the sum of the change in kinetic energy and the potential energy gained:
Total Energy Output = ΔKE + PE = 0.411 ft-lbs + 57.73 ft-lbs ≈ 58.14 ft-lbs
The mechanical energy lost during the climb can be calculated as:
Energy Lost = Work Done - Total Energy Output = 2000 ft-lbs - 58.14 ft-lbs ≈ 1941.86 ft-lbs
Finding the Normal Force
The normal force (N) acting on the block can be calculated by considering the forces acting perpendicular to the inclined plane. The weight of the block can be calculated as:
Weight = m × g = 3 kg × 32.2 ft/sec² = 96.6 lbs
Now, the component of the weight acting perpendicular to the incline is:
Weight_perpendicular = Weight × cos(θ) = 96.6 lbs × cos(60°) = 96.6 lbs × 0.5 = 48.3 lbs
Since there is no vertical acceleration, the normal force is equal to this perpendicular component of the weight:
Normal Force (N) = Weight_perpendicular = 48.3 lbs
Summary of Results
- Work Done by the 200 lbs Force: 2000 ft-lbs
- Mechanical Energy Lost on the Climb: Approximately 1941.86 ft-lbs
- Normal Force: 48.3 lbs
This breakdown illustrates how to approach problems involving forces, energy, and motion on an inclined plane. Each step builds on the previous one, leading to a comprehensive understanding of the mechanics involved.
