energy stored in spring= 1/2 k d^2
energy in f riction= mg*mu*.05m
leftover energy= 1/2 m 1.5^2
well, initial energy=finalenergy+frictionwork
solve for d
In Figure 8-34, a 1.2 kg block is held at rest against a spring with a force constant k = 695 N/m. Initially, the spring is compressed a distance d. When the block is released, it slides across a surface that is frictionless, except for a rough patch of width 5.0 cm that has a coefficient of kinetic friction µk = 0.44. Find d such that the block's speed after crossing the rough patch is 1.5 m/s.
2 answers
energy in f riction= mg*mu*.05m = 5.886
leftover energy= 1/2 m 1.5^2 = 1.35
So, If I add the leftover energy and the energy in friction I get the initial energy.
Then what??
leftover energy= 1/2 m 1.5^2 = 1.35
So, If I add the leftover energy and the energy in friction I get the initial energy.
Then what??