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Work, Energy and Power

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Physics Part I

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Physics

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.


Given,

Mass of the block, 
m = 1 kg 

Spring constant, k = 100 N m–1 

Displacement in the block, x = 10 cm = 0.1 m

The given situation can be illustrated in the following figure, 



At equilibrium, 

Normal reaction, R = mg cos 37° 

Frictional force, f = μ R

                          = mg Sin 37

where, μ is the coefficient of friction

Net force acting on the block = mg sin 37° – 

                                            = mgsin 37° – μmgcos 37° 

                                            = mg(sin 37° – μcos 37°) 

At equilibrium, the work done by the block is equal to the potential energy of the spring.

i.e., 
mg(sin 37° – μcos 37°)x = open parentheses 1 half close parentheses kx2
 1 × 9.8 (Sin 370 - μcos 37°) =open parentheses 1 half close parentheses × 100 × (0.1) 

                0.602 - μ × 0.799 = 0.510 

∴                        μ = fraction numerator 0.092 over denominator 0.799 end fraction  =  0.115 





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