Answered: A 1.80 kg block slides 3 meters on a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Chapter 7 Physics: Use conservation of Energy

A 1.80 kg block slides 3 meters on a rough horizontal surface before hitting a spring.
When the block hits the spring, it compresses the spring a distance of 15 cm before
coming to rest. If the coefficient of kinetic friction between the block and the surface
is 0.50 and the spring constant is 250 N/m, what was the speed of the block at the
beginning of the 3 meters?

Expert Solution

Step 1

Given Data:-

Mass of the block is $m = 1.80 k g$ .

Spring constant of the spring is $k = 250 N / m$ .

Distance travelled by the block through the rough surface is $d = 3.0 m$ .

Coefficient of kinetic friction between the surface and the block is $μ_{k} = 0.50$ .

Compression of the spring is

$∆ x = 15 c m$

$\Rightarrow ∆ x = \frac{15}{100} m$

$\Rightarrow ∆ x = 0.15 m$

Explanation:-

If initial speed of the block is $v_{i}$ then its initial kinetic energy is $K_{i} = \frac{1}{2} m {v_{i}}^{2}$ .

Final kinetic energy of the block is $E_{f} = 0$

Work done against frictional force is

$W_{f} = - μ_{k} m g d$ [here negative sign implies that frictional fore is acting opposite to the displacement of the block.]

If the spring compressed by a length $∆ x$ then potential energy stored in the spring is $E_{p} = \frac{1}{2} k ∆ x^{2}$ .

Now from the conservation of mechanical energy we get,

$K_{f} - K_{i} + E_{p} = W_{f}$

$\Rightarrow 0 - \frac{1}{2} m {v_{i}}^{2} + \frac{1}{2} k ∆ x^{2} = - μ_{k} m g d$

$\Rightarrow \frac{1}{2} m {v_{i}}^{2} - \frac{1}{2} k ∆ x^{2} = μ_{k} m g d$

$\Rightarrow \frac{1}{2} m {v_{i}}^{2} = (\frac{1}{2} k ∆ x^{2} + μ_{k} m g d)$

$\Rightarrow \frac{1}{2} m {v_{i}}^{2} = \frac{k ∆ x^{2} + 2 μ_{k} m g d}{2}$

$\Rightarrow {v_{i}}^{2} = \frac{(k ∆ x^{2} + 2 μ_{k} m g d)}{m}$

$\Rightarrow v_{i} = \sqrt{\frac{(k ∆ x^{2} + 2 μ_{k} m g d)}{m}}$ .................(A)

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