03-06-2013, 08:12 AM |
#1 |

Junior Member Join Date: Mar 2013
Posts: 14
| Some questions I can't get. I can't figure out how to work these physics problems. Question 3 An applied force of 100 N causes a 20 kg crate to accelerate across a level floor at 2 m/s2. Draw a free body diagram illustrating this situation and find the coefficient of kinetic friction between the crate and the floor. I know that FN=FG=20*9,8=196N, and that the acceleration = FNet/Mass which means Fnet=2*20=40 but I don't know how to get the kenetic friction. Question 4 60 N of force is applied to a 10 kg crate initially at rest on a level floor. The coefficient of friction between the crate and the floor is 0.4. Draw a free body diagram illustrating this situation and find the distance traveled and the velocity of the crate after 3 seconds. again FN=FG=10*9.8=198N, but I don't know where to go from there. |

03-06-2013, 09:54 AM |
#2 |

Junior Member Join Date: Dec 2012
Posts: 5
| With Q3 the force of friction is 100 - ma =100 - 20*2=60N. The limiting value of the frictional force is muR, where mu is called the coefficient of friction and R is the normal reaction for the surfaces in contact. mu can be calculated to be 0.306 |

03-06-2013, 09:58 AM |
#3 |

Senior Member Join Date: Jun 2010
Posts: 1,945
| For #3 use the relationship Sum of Forces = ma. You found sum of forces already (you called it F_net), but what do those forces consist of? Theres' one force pushing on the box (196N) and another force resisting that movement. So F_friction = F_net - F_push. Then use the concept that F_friction = Normal force (i.e. weight) times coefficient of friction, and solve for the coefficient of friction. For #4 the forces acting on the crate are the 60N push minus the force of friction, which is its normal force to the floor (it's weight in this case) times the coefficient of friction. Set his sum of forces equal to ma, and solve for a. Now yuo can determine distance and velocity using standard equations of motion, such as v=at and d = (1/2)at^2. |

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