Chapter 9: Q.70 (page 230)

A $12 k g$ weather rocket generates a thrust of $200 N .$ The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is $550 N / m$ , is anchored to the ground.
a. Initially, before the engine is ignited, the rocket sits at rest on top of the spring. How much is the spring compressed?
b. After the engine is ignited, what is the rocket’s speed when the spring has stretched $40 c m ?$

Short Answer

Expert verified

a. The compression in the spring is $21 c m .$

b. The speed of rocket is $1.73 m / s$

Step by step solution

Content Introduction

The conservation of energy is

$K_{i} + U_{i} + W = K_{f} + U_{f} + ∆ E_{t h}$

Considering the following diagram,

The origin of the coordinate system is placed on a free ed of the spring.

Explanation (Part a)

The rocket is initially at rest. Apply the newton's second law to balance the forces.

$\sum_{} F_{y} = 0 F_{s p} - F_{g} = 0 F_{s p} = F_{g} k ∆ y = m g$

Therefore,

$∆ y = \frac{m g}{k} ∆ y = \frac{12 k g (9.8 m / s^{2})}{550 N / m} ∆ y = 21 c m$

Explanation (Part b)

Use the equation for conservative energy when $y_{2} - y_{e} = 40 c m K_{2} + U_{g, 2} + U_{s p, 2} = K_{1} + U_{g, 1} + U_{s p, 1} + W_{e x t} \frac{1}{2} m {v_{2}}^{2} + m g y_{2} + \frac{1}{2} k {(y_{2} - y_{e})}^{2} = \frac{1}{2} m v$ ₁2+mgy1+12k(y1-ye)212(12kg)v22+(12kg)(9.8m/s2)(40cm.1m100cm)+12(550N/m)(40cm.1m100cm)12(12kg)v22+47J+44J=-24J+11J+122Jv2=18J6kgv2=1.73m/s