A 20.0 kg box slides down a 12.0 m long incline at an angle of 3.0 degrees with the horizontal. A force of 50.0 N is applied to the box to try to pull it up the incline. The applied force makes an angle of 0.00 degrees to the incline. If the incline has a coefficient of kinetic friction of 0.100, then the increase in the kinetic energy of the box is:
3 answers
A 20.0 kg box slides down a 12.0 m long incline at an angle of 3.0 degrees with the horizontal. A force of 50.0 N is applied to the box to try to pull it up the incline. The applied force makes an angle of 0.00 degrees to the incline. If the incline has a coefficient of kinetic friction of 0.100, then the increase in the kinetic energy of the box is:
ok,
vf^2=2ad
acceleration=50/mass-20g*cos3.0/mass-.1*g*sin3.0
solve for vf
then, KE increase=1/2 m vf^2
vf^2=2ad
acceleration=50/mass-20g*cos3.0/mass-.1*g*sin3.0
solve for vf
then, KE increase=1/2 m vf^2
300