A 6 kg block free to move on a horizontal, frictionless surface is attached to a spring. The spring is compressed 0.16 m from equi- librium and released. The speed of the block is 1.44 m/s when it passes the equilibrium po- sition of the spring. The same experiment is now repeated with the frictionless surface re- placed by a surface with coefficient of friction 0.1.
Determine the speed of the block at the equilibrium position of the spring. The accel- eration due to gravity is 9.8 m/s2 .
2 answers
Someone please answer i need help and do not understand
kinetic energy of block released from spring = (1/2) mv^2 = .5*6 (1.44)^2
= 6.22 Joules
Now if there is friction, how much work is done against friction force?
Work = F * d
work = .1 * 6 * 9.8 * .16 = .941 Joules lost to friction
so kinetic energy left in block = 6.22 -.941 = 5.28 Joules
so
(1/2) m v^2 = 5.28 Joules
v^2 = (2 /6)(5.28)
v = 1.33 m/s
so energy stored in spring = 6.22 Joules if no friction
(1/2) k x^2 = 6.22
(1/2) k (.16)^2 = 6.22
so
spring constant k = 486 N/m
= 6.22 Joules
Now if there is friction, how much work is done against friction force?
Work = F * d
work = .1 * 6 * 9.8 * .16 = .941 Joules lost to friction
so kinetic energy left in block = 6.22 -.941 = 5.28 Joules
so
(1/2) m v^2 = 5.28 Joules
v^2 = (2 /6)(5.28)
v = 1.33 m/s
so energy stored in spring = 6.22 Joules if no friction
(1/2) k x^2 = 6.22
(1/2) k (.16)^2 = 6.22
so
spring constant k = 486 N/m
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