1. Neither. Pushing down does nothing, pulling up moves it vertically.
2. Amazing Kreskin abilities fading...
3. The angle is 23.6o so Normal force is 10*9.8*cos23.6 and the friction force is .3 * Normal force. Friction force will give you work lost to friction (Fd where d =15). Finally take initial PE = mgh minus work lost to friction and set it equal to final KE (.5 m v^2).
1. A 50.0 Lb box is to be moved along a rough floor at a constant velocity. The coefficient of the kinetic friction is 0.300.
A. What force must you exert if you push downward on the box?
B. What force must you exert if you pull upward on the box?
C. Which is the better way to move the box?
2. Find :
A. The magnitude of the acceleration of the system shown if the coefficient of kinetic friction B is 0.300, Coefficient if kinetic friction A is 0.200, MB=3.00 kg, MA= 5.00kg.
B. The velocity of block A at 0.500s?
3. A 10.0 kg package slides down on inclined mail chute 15.0m long. The top of the chute is 6.00 above the floor. What is the speed of package at the button of the chute if the coefficient of kinetic friction is 0.300?
1 answer
1 answer