---
title: "Block sliding on incline compressing spring"
subject: "Engineering Mechanics"
chapter: 3
key: "semester_3/engineering_mechanics:block-incline-spring-compression"
source: https://www.bctengineer.com/q/semester_3/engineering_mechanics/block-incline-spring-compression
site: BCT Engineering
---

# Block sliding on incline compressing spring

Engineering Mechanics · Chapter 3: Kinetics of Particles: Work Energy Principles

- Asked in: year not given, 8 marks
- Video: [Dynamics: Lesson 24 - Work of Springs and Friction](https://www.youtube.com/watch?v=TVGUVPoYmfE) — Jeff Hanson
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/engineering_mechanics/block-incline-spring-compression
- Whole chapter: https://www.bctengineer.com/md/semester_3/engineering_mechanics/chapter-3

## Question

A 10 kg block is released from rest on a rough incline of 30° to the horizontal (coefficient of kinetic friction 0.20). After sliding 4 m along the incline it strikes a spring of stiffness 2000 N/m lying along the incline, whose axis is parallel to the incline. Using the work-energy principle, find the maximum compression of the spring and the speed of the block just before it touches the spring.

## Short answer

Net force down incline: mg sin30° – μ mg cos30° = 32.06 N. Speed before spring: v = √(2 F_net d / m) = 5.06 m/s. Work-energy over (4+x): F_net (4+x) – ½ k x² = 0, solving gives x = 0.374 m.

## Answer

**Given:** $m = 10$ kg, $\theta = 30^\circ$, $\mu = 0.20$, $d = 4$ m, $k = 2000$ N/m.

```text
   \
    \  block
     \[#]
      \ \  4 m
       \ \_____
        \  /\/\/\ spring
         \/_______
```

**Forces along the incline**
- Component of weight down the slope: $mg\sin30^\circ = 10 \times 9.81 \times 0.5 = 49.05$ N
- Normal reaction: $N = mg\cos30^\circ = 84.96$ N
- Friction (up the slope): $\mu N = 0.20 \times 84.96 = 16.99$ N
- Net driving force: $49.05 - 16.99 = 32.06$ N

#### Speed just before touching the spring
Work-energy over 4 m: $0 + F_{net}\, d = \tfrac12 m v^2$

$$
v = \sqrt{\frac{2 \times 32.06 \times 4}{10}} = 5.06\ \text{m/s}
$$

#### Maximum compression x
The block moves $(4 + x)$ along the incline from rest to rest (momentarily stopped at maximum compression). The spring does negative work $-\tfrac12 kx^2$.

$$
F_{net}(4 + x) - \tfrac12 k x^2 = 0
$$

$$
1000x^2 - 32.06x - 128.23 = 0
$$

$$
x = \frac{32.06 + \sqrt{32.06^2 + 4(1000)(128.23)}}{2000} = 0.374\ \text{m}
$$

(The negative root is rejected.)

**Answer: Maximum compression = 374 mm (0.374 m); speed before touching the spring = 5.06 m/s.**

## Check yourself

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

1. What is the net force down the incline on the 10 kg block?
   A. mg sin30 = 49.05 N
   B. mg sin30 - mu mg cos30 = 32.06 N
   C. mg sin30 + mu mg cos30 = 66.04 N
   D. mu mg cos30 = 16.99 N
   Answer: B. Friction 0.2 x 84.96 = 16.99 N acts up the slope: 49.05 - 16.99 = 32.06 N. (Key point: Net driving force)

2. Which equation gives the speed after 4 m?
   A. F_net x 4 = (1/2) m v^2
   B. mg x 4 = (1/2) m v^2
   C. F_net x 4 = m v
   D. (F_net + 16.99) x 4 = (1/2) m v^2
   Answer: A. Starting from rest, the work of the net force over 4 m equals the kinetic energy, giving v = 5.06 m/s. (Key point: Speed before spring)

3. In the maximum compression analysis, the net force acts over what distance?
   A. x only
   B. 4 m only
   C. (4 + x)
   D. (4 - x)
   Answer: C. The block slides 4 m to reach the spring and a further x while compressing it. (Key point: Distance used for compression)

4. Which equation determines x?
   A. F_net(4 + x) + (1/2)k x^2 = 0
   B. F_net(4 + x) - (1/2)k x^2 = 0
   C. F_net x - (1/2)k x^2 = 0
   D. F_net(4) = (1/2)k x^2
   Answer: B. At maximum compression the block is momentarily at rest, so net-force work minus spring work is zero. Solving gives x = 0.374 m. (Key point: Spring work)

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