Question

A block of mass m=1kg sliding along a rough horizontal surface is traveling at a speed v0=2m/s when it strikes a massless spring head-on and compresses the spring a maximum distance X=10cm. If the spring has stiffness constant k=10N/m, determine the coefficient of kinetic friction between block and surface.

Answer 1

Here we can use the work energy theorem

$W_f + W_s = K_f - K_i$

here we know that

$K_f = 0$

as it come to rest finally

$K_i = \frac{1}{2}mv_i^2$

$K_i = \frac{1}{2}\times 1\times 2^2$

$K_i = 2 J$

now work done by friction force will be given as

$W_f = - F_f \times d = -\mu mg d$

$W_f = - \mu(1)(9.8)(0.10) = - 0.98\mu$

Work done by spring force is given as

$W_s = \frac{1}{2}k(x_i^2 - x_f^2)$

$W_s = \frac{1}{2}(10)( 0 - 0.10^2)$

$W_s = -0.05 J$

so now plug in all data above

$- 0.05 - \mu(0.98) = 0 - 2$

$\mu = 1.99$

so above is the friction coefficient

Answer 2

The coefficient of kinetic friction between block and surface is :  1.99

Given data :

mass of block ( m ) = 1 kg

Speed of block ( $v_{i}$ ) = 2 m/s

spring compression ( x ) = 10 cm = 0.1 m

stiffness constant ( k ) = 10 N/m

Determine the value of coefficient of kinetic friction between block and surface

Applying the work energy theorem

Wf + Ws = Kf - Ki  ------ ( 1 )

where : Kf = 0,  Ki = $1/2 * mv^{2} _{i}$ = 2J ,  

Back to equation ( 1 )

Work done by frictional force ( Wf ) becomes

           Wf   = - μmgd

                   =  - μ * 1 * 9.8 * 0.10

                   = -0.98 μ

work done by Ws

Ws = $1/2k(x^2_{i} - x^2_{f} )$

     = 1/2 * 10 ( 0 - 0.10² )

     = -0.05 J

Insert all values into equation ( 1 )

μ ( coefficient of kinetic friction between block and surface ) = 1.99

Hence we can conclude that The coefficient of kinetic friction between block and surface is :  1.99

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