Question

The spring in FIGURE EX10.22a is compressed by ∆x. It launches the block across a frictionless surface with speed v0. The two springs in FIGURE EX10.22b are identical to the spring of Figure EX10.22a. They are compressed by the same ∆x and used to launch the same block. What is the block’s speed now?

Short answer

The speed of the block will now be two times faster than when it was launched with a single spring$v=2v0\mathrm{vV}0=2.$

Step-by-step solution

Introduction

This implies that the speed and net force in the plane's perpendicular direction must be zero. Assuming that the plane is "frictionless," this means that the plane exerts no force on the parallel-to-the-surface block.

Explanation (a)

when just one springs

we can say that $\mathrm{k}$be the spring constant

compression of the spring is$=\Delta xm$which is mass of block

we see that potential energy of spring is = kinetic energy of block

$12\mathrm{k}\left(\Delta \mathrm{x}\right)2=12\mathrm{mv}02\mathrm{v}0=\Delta \mathrm{xkm}$

Explanation (b)

When two springs are present

The potential energy of spring is equal to the kinetic energy of block

spring's potential energy can be found here$=2\times 12k\left(\Delta x\right)2=k\left(\Delta x\right)2$

$\mathrm{k}\Delta \mathrm{x}2=12\mathrm{mv}2\mathrm{v}$is speed $=\Delta x2\mathrm{km}=\Delta \mathrm{xkm}\times 2=2\mathrm{v}0$.

The speed of the block will now be two times faster than when it was launched with a single spring

$v=2v0\mathrm{vV}0=2$.