Diagram of a particle on an inclined plane with applied force and force labels.

Let’s solve the first problem shown in the image — the one with the 200kg particle on a 40° slope, with a 1500N force applied up the slope.

We are to:
- Label all forces acting on the particle (we’ll describe them since we can’t draw here).
- Calculate the acceleration of the particle.

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Step 1: Understand the situation

A 200 kg block is on a slope inclined at 40° to horizontal.
A force of 1500 N is applied up the slope (parallel to it).
We assume no friction unless stated — and this problem doesn’t mention friction, so we’ll assume no friction.

Forces acting on the particle:

1. Weight (W) = mass × gravity = 200 kg × 9.8 m/s² = 1960 N → acts straight down.
2. Normal force (N) → perpendicular to the slope, pushing the block away from the surface.
3. Applied force (F_app) = 1500 N → up the slope.
4. Component of weight parallel to slope → pulls the block *down* the slope.
5. Component of weight perpendicular to slope → balanced by normal force.

Since motion is along the slope, we only care about forces parallel to the slope for acceleration.

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Step 2: Break weight into components

Weight has two parts relative to the slope:

- Parallel component (pulls block down slope):
W_parallel = mg sin(θ) = 200 × 9.8 × sin(40°)

- Perpendicular component (pressed into slope):
W_perp = mg cos(θ) → but this doesn’t affect motion along slope if no friction.

Calculate W_parallel:

sin(40°) ≈ 0.6428

So,

W_parallel = 200 × 9.8 × 0.6428
= 1960 × 0.6428
≈ 1259.89 N (down the slope)

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Step 3: Net force along the slope

Applied force is up the slope: +1500 N
Weight component is down the slope: -1259.89 N

Net force F_net = 1500 - 1259.89 = 240.11 N (up the slope)

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Step 4: Use Newton’s Second Law to find acceleration

F_net = m × a
→ a = F_net / m = 240.11 / 200 ≈ 1.20055 m/s²

Rounded reasonably: 1.20 m/s²

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Final Answer:

The acceleration of the particle is 1.20 m/s² up the slope.