---
title: "30 kg block on spring scale, k = 20 kN/m"
subject: "Applied Mechanics (Dynamics)"
chapter: 2
key: "semester_3/applied_mechanics_dynamics:pan-spring-scale-30kg-k20"
source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/pan-spring-scale-30kg-k20
site: BCT Engineering
---

# 30 kg block on spring scale, k = 20 kN/m

Applied Mechanics (Dynamics) · Chapter 2: Kinetics of particles: Energy and Momentum Methods

- Asked in: 2073 Shrawan, 8 marks
- Video: [Dynamics - Finding maximum deflection of the block when blockbis released from 2m height Applied](https://www.youtube.com/watch?v=e22vTiYewTk) — Sudip Dhakal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/pan-spring-scale-30kg-k20
- Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2

## Question

A 30 kg block is dropped from a height of 2 m onto the 10 kg pan of a spring scale. Assuming the impact to be perfectly plastic, determine the maximum deflection of the pan. The constant of the spring is $k = 20$ kN/m. [Figure: 30 kg block A above 10 kg pan B on a spring, drop height 2 m.]

## Short answer

The block strikes at 6.264 m/s; after plastic impact common speed is 4.698 m/s. Initial spring compression from pan weight is 0.0049 m. Work-energy yields total compression 0.230 m, so pan deflection ≈0.225 m below impact position.

## Answer

**Data:** $m_A=30$ kg, $m_B=10$ kg, $h=2$ m, $k=20$ kN/m, perfectly plastic impact ($e=0$), $g=9.81\ \text{m/s}^2$.

#### Step 1: block just before impact
$$
v_A=\sqrt{2gh}=\sqrt{2(9.81)(2)}=6.264\ \text{m/s}
$$

#### Step 2: plastic impact (pan initially at rest)
Momentum is conserved and both move together:

$$
30(6.264)=(30+10)v'\ \Rightarrow\ v'=4.698\ \text{m/s}
$$

#### Step 3: initial compression of the spring due to the pan
$$
x_1=\frac{m_Bg}{k}=\frac{10(9.81)}{20000}=0.00490\ \text{m}
$$

#### Step 4: work-energy from just after impact to the lowest point
Let $x_2$ be the maximum total compression and $M=40$ kg. Kinetic energy just after impact: $\tfrac12Mv'^2=441.45$ J.

$$
\tfrac12Mv'^{2}+\tfrac12kx_1^{2}+Mg(x_2-x_1)=\tfrac12kx_2^{2}
$$

$$
10000x_2^{2}-392.40x_2-\left(441.45+0.2406-1.9247\right)=0
$$

$$
x_2=0.2302\ \text{m}
$$

The pan moves down $x_2-x_1=0.2253$ m from its position at impact.

**Answer: maximum deflection of the pan $\approx0.225$ m below its position at impact (total spring compression $\approx0.230$ m).**

## Check yourself

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

1. How is the block's speed just before hitting the pan found?
   A. v = sqrt(2gh)
   B. v = g h
   C. v = sqrt(g h)
   D. v = 2 g h
   Answer: A. Free fall from rest through h gives v = sqrt(2gh) = 6.264 m/s. (Key point: Block speed before impact)

2. Which principle gives the common speed after the perfectly plastic impact?
   A. Conservation of energy
   B. Conservation of linear momentum
   C. Work-energy of the spring
   D. Newton's third law only
   Answer: B. Impact is very short; momentum is conserved, giving v' = 4.698 m/s, but energy is lost. (Key point: Impact stage)

3. What gives the spring's initial compression x1 under the pan alone?
   A. x1 = m_B g / k = 0.00490 m
   B. x1 = (m_A + m_B) g / k
   C. x1 = 0
   D. x1 = k / (m_B g)
   Answer: A. The pan (10 kg) alone is in equilibrium on the spring before impact. (Key point: Initial compression)

4. What does the answer 0.225 m represent?
   A. Total spring compression x2
   B. Downward movement of the pan from its position at impact, x2 - x1
   C. Height fallen by the block
   D. Initial compression x1
   Answer: B. Total compression x2 = 0.230 m, so the pan moves x2 - x1 = 0.225 m below its position at impact. (Key point: Final answer meaning)

## Close variants set in other exams

- [30 kg block on spring scale, k = 30 kN/m](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Apan-spring-scale-30kg-k30) — Applied Mechanics (Dynamics), chapter 2, asked 3×
- [40 kg block on spring scale, central impact](https://www.bctengineer.com/question.md?key=semester_3%2Fapplied_mechanics_dynamics%3Apan-spring-scale-40kg) — Applied Mechanics (Dynamics), chapter 2, asked 3×