Answered: No friction D (H)

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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Question

A spring is initially compressed and has a spring constant of 1865 N/m. When the spring is released, it pushes on a 4 kg block causing the block to accelerate. There is no friction between the table and block during the acceleration. The block leaves the spring at the spring’s relaxed length. Starting from this point, there is friction between the block and table. The block slides along the table with a coefficient of kinetic friction µk = .25. The block stops in a distance D = 2.4 m. What are: a. the increase in the thermal energy? b. the maximum kinetic energy of the block? c. the original compression distance of the spring?

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Expert Solution

Step 1

the spring constant K=1865N/m

The block will stop at a distance D=2.4m

$t h e i n i t i a l v e l o c i t y w i t h w h i c h i t s t a r t e d = u t h e f i n a l v e l o c i t y = v = 0 m / s - a = r e t a r d i n g a c c e l e r a t i o n p r o v i d e d b y t h e f r i c t i o n a l f o r c e = μ g u \sin g t h e a b o v e d a t a i n e q u a t i o n o f m o t i o n v^{2} = u^{2} - 2 a s 0 = u^{2} - 2 μ g D u^{2} = 2 μ g D u = \sqrt{2 μ g D} = \sqrt{2 \times 0.25 \times 9.8 \times 2.4} m / s = 3.43 m / s$

Step 2

$t h e f o r c e a p p l i e d b y t h e s p r i n g = k x w h i c h w i l l a c c e l e r a t e t h e b l o c k t o 3.43 m / s f r o m 0 m / s s o a g a i n u \sin g t h e e q u a t i o n o f m o t i o n v^{2} = u^{2} + 2 a x i n t h i s c a s e u = 0 m / s a n d v = 3.43 m / s i t c a n a l s o b e g i v e n t h a t m a = k x a = \frac{k x}{m} {(3.43)}^{2} = 0 + 2 a x {(3.43)}^{2} = 2 x (\frac{k x}{m}) x^{2} = \frac{{(3.43)}^{2} \times m}{2 k} = \frac{{(3.43)}^{2} \times 4}{2 \times 1865} = 0.0126 x = 0.112 m (c o m p r e s s i o n i n t h e s p r i n g)$

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