Answered: I need help with 5a ,5b and 6a also the…

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Question

I need help with 5a ,5b and 6a

also the final numbers should be rounded to tree significant digits

thanks

Expert Solution

Step 1 Concept

The problem is based on the concept of Kinetic friction.

Diagram:

(5a) Given;

$\begin{array}{rcl} M & = & 81 \cdot 1 k g \\ g & = & 9 \cdot 8 m / s^{2} \\ F & = & 284 N \\ = & T h i s i s f o r c e a p p l i e d o n t h e c l o c k t o s l i d e o n t h e f l o o r \\ = & T h i s f o r c e i s o p p o s i t e t o f r i c t i o n a l f o r c e f_{k} . \\ N o w t h e m a g n i t u d e o f t h e k i n e t i c f r i c t i o n f o r c e a c t i n g o n t h e c l o c k i s; \\ f_{k} & = & μ_{k} N \\ W h e r e; \\ f_{k} & = & T h e m a g n i t u d e o f t h e k i n e t i c f r i c t i o n f o r c e \\ μ_{k} & = & T h e c o e f f i c i e n t o f k i n e t i c f r i c t i o n \\ N & = & T h e n o r m a l f o r c e \\ f_{k} & = & μ_{k} N \\ = & μ_{k} \times (M g) \\ f_{k} & = & μ_{k} 794 \cdot 78 N \\ A n s w e r : T h e m a g n i t u d e o f t h e k i n e t i c f r i c t i o n f o r c e a c t i n g o n t h e c l o c k i s μ_{k} 794 \cdot 78 N \end{array}$

(5b) As given in the problem,

$\begin{array}{rcl} T h e h o r i z o n t a l f o c e a p p l i e d o n t h e c l o c k t o t o s l i d e o n t h e f l o o r i s F . \\ F & = & 284 N \\ f_{k} & = & K i n e t i c f r i c t i o n f o r c e = μ_{k} M g \\ A s w e k n o w t h e h o r i z o n t a l f o r c e a n d k i n e t i c f r i c t i o n f r o c e a r e o p p o s i t e t o e a c h o t h e r . \\ a c c o r d i n g t o N e w t o n' s s e c o n d l a w o f m o t i o n; \\ M a & = & F - μ_{k} M g \\ b u t a s g i v e n i n t h e p r o b l e m (5 a) c l o c k s l i d e s a c r o s s t h e f l o o r w i t h c o n s \tan t v e l o c i t y . \\ T h e r e f o r e, \frac{d v}{d t} & = & a = 0 \\ T h e a b o v e e q u a t i o n c a n b e w r i t t e n a s; \\ F - μ_{k} M g & = & 0 \\ μ_{k} & = & \frac{F}{M g} \\ = & \frac{284}{(81 \cdot 1 \times 9 \cdot 8)} \\ = & \frac{284}{794 \cdot 78} \\ = & 0 \cdot 357 \\ ∴ μ_{k} & = & 0 \cdot 357 \\ A n s w e r (5 b) & = & T h e c o e f f i c i e n t o f k i n e t i c f r i c t i o n b e t w e e n t h e c l o c k a n d t h e f l o o r i s μ_{k} = 0 \cdot 357 \\ Therefore the magnitude of kinetic friction force acting on the clock calculated in problem (5 a) can be modified as \\ f_{k} & = & μ_{k} 794 \cdot 78 \\ = & (0 \cdot 357 \times 794 \cdot 78) \\ = & 283 \cdot 736 N \\ A n s w e r (5 a) & = & T h e m a g n i t u d e o f k i n e t i c f r i c t i o n f o r c e a c t i n g o n t h e c l o c k i s 283 \cdot 736 N \end{array}$

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