---
title: "Force P to give block acceleration of 10 m/s²"
subject: "Applied Mechanics"
chapter: 9
key: "semester_1/applied_mechanics:force-p-block-2500n-acceleration-10"
source: https://www.bctengineer.com/q/semester_1/applied_mechanics/force-p-block-2500n-acceleration-10
site: BCT Engineering
---

# Force P to give block acceleration of 10 m/s²

Applied Mechanics · Chapter 9: Kinetics of Particles and Rigid Body: Force and Acceleration

- Asked in: 2070 Asar, 10 marks
- Video: [Force Required to Produce Certain Acceleration on Frictional Surface](https://www.youtube.com/watch?v=rb9P3IGMjgk) — Indra Lal Subedi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/force-p-block-2500n-acceleration-10
- Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-9

## Question

Determine the magnitude of force P required to give the block an acceleration of 10 m/s². Coefficient of friction between the block and the floor is 0.25. [Figure: block of weight W = 2500 N on the floor pulled by force P acting at 30° above the horizontal.]

## Short answer

Resolving forces vertically: N = 2500 - 0.5P. Horizontal equation: P cos30° - 0.25N = (2500/9.81)×10. Solving yields 0.991P = 3173.4, so P = 3202 N.

## Answer

#### Reading of the figure
The block (W = 2500 N) rests on a horizontal floor and is pulled by P acting at 30° above the horizontal (the arrow points up and to the right). μ = 0.25, required acceleration a = 10 m/s².

```text
          P  /
            / 30°
   [ W=2500 N ]--->  a = 10 m/s²
 ---------------------------
```

#### Equations of motion
- Vertical (no vertical motion): N + P sin 30° − 2500 = 0, so N = 2500 − 0.5P
- Horizontal: P cos 30° − μN = m a, with m = 2500/9.81 = 254.84 kg

P(0.8660) − 0.25(2500 − 0.5P) = 254.84 × 10
0.8660P − 625 + 0.125P = 2548.4
0.9910P = 3173.4

**P = 3202 N ≈ 3.2 kN**

Check: N = 2500 − 1601 = 899 N, friction = 224.8 N, P cos 30° = 2773 N, net = 2548 N = m a ✓.

**Answer: P ≈ 3202 N.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. P acts 30 degrees above the horizontal. What is N?
   A. N = 2500 + P sin 30
   B. N = 2500
   C. N = 2500 - P sin 30
   D. N = 2500 - P cos 30
   Answer: C. The vertical component P sin 30 lifts part of the weight, so N = 2500 - 0.5P. (Key point: Normal reaction with upward P)

2. What mass is used in m a?
   A. 2500 kg
   B. 2500/9.81 = 254.84 kg
   C. 2500 x 9.81
   D. 250 kg exactly
   Answer: B. Weight is given, so m = W/g = 254.84 kg and m a = 2548 N. (Key point: Mass from weight)

3. Which horizontal equation leads to P = 3202 N?
   A. P sin 30 - mu N = m a
   B. P cos 30 + mu N = m a
   C. P cos 30 - mu(2500 - 0.5P) = 254.84 x 10
   D. P = m a
   Answer: C. Friction mu N opposes motion, giving 0.866P - 625 + 0.125P = 2548.4, so P = 3202 N. (Key point: Solving for P)

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