Thank you for the opportunity to help you with your question!
(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.
Content will be erased after question is completed.