SOLUTION: The total energy required to move a 10-N block up a 25-degree incline for 10 meters is 500 Joule. A frictional force of 10-N opposes the motion of the block. Determine force (paral

Question 1182648: The total energy required to move a 10-N block up a 25-degree incline for 10 meters is 500 Joule. A frictional force of 10-N opposes the motion of the block. Determine force (parallel the inclined) required to move the block, the rate of work done if the work was done in 10 seconds, and the initial velocity of the block if its velocity at 10 meters is 30 m/s.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's the solution:
**1. Force parallel to the incline:**
* **Work done against gravity (W_g):**
W_g = mgh = mg * d * sin(θ)
Where:
* m = mass of the block (weight = mg = 10 N, so m = 10 N / g, where g ≈ 9.8 m/s²)
* h = vertical height gained = d * sin(θ)
* d = distance moved along the incline = 10 m
* θ = angle of incline = 25°
W_g = (10 N / 9.8 m/s²) * 9.8 m/s² * 10 m * sin(25°)
W_g ≈ 42.3 J
* **Work done against friction (W_f):**
W_f = frictional force * distance
W_f = 10 N * 10 m = 100 J
* **Total work done (W_total):** This is given as 500 J.
* **Work done by the applied force (W_applied):**
W_total = W_g + W_f + W_applied
500 J = 42.3 J + 100 J + W_applied
W_applied ≈ 357.7 J
* **Force parallel to the incline (F):**
W_applied = F * d
357.7 J = F * 10 m
F ≈ 35.77 N
**2. Rate of work done (Power):**
Power = Work done / Time
Power = 500 J / 10 s = 50 W
**3. Initial velocity:**
We can use the work-energy theorem to find the initial velocity. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy.
* **Change in kinetic energy (ΔKE):**
ΔKE = (1/2)mv² - (1/2)mv₀²
Where:
* v = final velocity = 30 m/s
* v₀ = initial velocity (what we want to find)
* **Net work done (W_net):** The net work is the total work minus the work done against friction.
W_net = W_total - W_f = 500J - 100J = 400 J
* **Apply the work-energy theorem:**
W_net = ΔKE
400 J = (1/2) * (10 N / 9.8 m/s²) * (30 m/s)² - (1/2) * (10 N / 9.8 m/s²) * v₀²
400 J ≈ 459.2 J - 0.51v₀²
0.51v₀² ≈ 59.2 J
v₀² ≈ 116.1
v₀ ≈ √116.1 ≈ 10.77 m/s
**Summary:**
* Force parallel to the incline: ≈ 35.77 N
* Rate of work done: 50 W
* Initial velocity: ≈ 10.77 m/s

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