Physics problem to solve

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A 200 g block is pressed against a spring of force constant 1.40 kN/m until the block compresses the spring 10.0 cm. The spring rests at the bottom of a ramp inclined at 60◦ to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops (a) if the ramp exerts no friction force on the block and (b) if the coefficient of kinetic friction is 0.400.

Oct 24th, 2015

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(a)initial potential energy of spring = final gravitational potential energy of block 

The potential energy of a spring is: U = ½kx2

Final gravitational potential energy of the block

Workby gravity = mg(h)

 ½kx2= mgh Thus  since we have the angle  the potential  energy   equlas to  the gravitational energy  on the block

kx^2 / 2 = mgLsinθ 

L = kx^2 / [2mgsinθ] 

L = (1400 N/m)(0.100 m)^2 / [2(0.200 kg)(9.81 m/s^2)(sin60.0°)] 

L = 4.12 m 


Here there  is  an extra  work  done of the friction thus 

initial potential energy of spring = (work of friction) + (final gravitational potential energy of block) 

kx^2 / 2 = μNL + mgh

kx^2 / 2 = μmgLcosθ + mgLsinθ 

L = kx^2 / [2mg(μ*cosθ + sinθ)] 

L = (1400 N/m)(0.100 m)^2 / [2(0.200 kg)(9.81 m/s^2)((0.400)*cos60.0° + sin60.0°)] 

L = 3.35 m

Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 24th, 2015

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