parallel to the incline. If the coefficient of kinetic friction between box and incline is 0.220, calculate the change in the kinetic energy of the box.
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Need help on this one. Thanks guys
2007-02-26 15:20:17 · 1 answers · asked by cosmo 1 in Science & Mathematics ➔ Physics
f=uN
where f is the force of kinetic friction, u the coefficient of kinetic friction, and N is the normal to the surface.
the normal to surface on account of the weight of the box=83cos30. Substituting in the above formula:
f=0.220*83cos30
=15.8N
The force parallel to the incline=95N
The unbalanced force, F=95-15.8
=79.2N
F=ma
m=w/g
=83/9.8
=8.47kg
Substituting known values:
79.2=8.47a
a=79.2/8.47
=9.35m/s^2
v^2-u^2=2as
where v is the final velocity, u the initial velocity, a the acceleration, and s the displacement
Substituting known values:
v^2-0=2*9.35*24
v^2=448.8
Kinetic Energy=1/2 mv^2
Change in KE=final KE-initial KE
=1/2*8.47*448.8-0
=1900.7joules
2007-02-27 00:56:20 · answer #1 · answered by tul b 3 · 0⤊ 0⤋