.. A 10.0 Kg block is released from point A in the figure below. The block trave

Question

..

A 10.0 Kg block is released from point A in the figure below. The block travels along a smooth frictionless track except for the rough 1 m long portion between points B and C. At the end of the track, the 10 kg block hits a spring of force constant (k) = 2000 N/m, and compresses the spring a distance of 20 cm from its equilibrium position before coming to rest momentarily What is the speed of the block at point B? What is the speed of the block at point C? What is the value of the coefficient of kinetic friction (|ak) between the block and the rough portion between points B and C?

Explanation / Answer

Let the Coefficient of Friction in the Surface BC is u

Therefore

Upto point B loss in Potential Energy = Gain in Kinetic Energy

Let v be the velocity at point B

Therefore

mgh = 0.5*m*v^2

Therefore

v = sqrt(2gh)

= sqrt(2*9.8*1)

= 4.427 m/sec



Therefore

Kinetic Energy at Point B = 0.5*10*4.427^2

= 98 J

Work Done against Friction = umgs

= u*10*9.8*1

= 98u

Therefore

Rest Energy is Converted into Elastic Potential Energy of the Spriong

Thereofore

98 - 98u = 0.5*k*x^2

98*(1-u) = 0.5*2000*0.20^2

Therefore

u = 0.5918

Let the Velocity at C is V



Kinetic Energy at C = Kinetic Energy at B - work Done against Friction

0.5*10*V^2 = 98 - 0.5918*10*9.8*1

V = 2.828 m/sec

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