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Published byMeagan Cain Modified over 9 years ago
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WORK AP 1 chapter 6: section 6.1-6.2
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WORK The product of the magnitude of the displacement times the component of the force PARALLEL to the direction of motion. Scalar but can be negative or positive (lose or gain) W=F”d =Fdcos Ө Ө is the angle between the force and the direction of motion. Units are Newton-meter (Nm)= Joule (J)
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WORK CAN BE Done by the object (-W) Done on the object (+W) Done by a specific force (must name it) Done by a Net force
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A PERSON PULLS A 50 KG CRATE 40 M ALONG A HORIZONTAL FLOOR BY A CONSTANT FORCE OF 100N ACTING AT A 37°ANGLE. THE FLOOR IS ROUGH AND EXERTS A FRICTIONAL FORCE OF 50N. Find the work done by each force. Wg done by gravity =mgdcos Ө = 0 WN done by FN =Fndcos Ө = 0 WA done by person =Fadcos Ө= 100N(40m)cos 37 =3200J Wf done by friction =Ffdcos Ө= 50N(40m)cos180 = -2000J FN Ff Fa Fg
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NET WORK ON CRATE Wnet= sum of all work =0J + 0J +3200J + -2000J =1200 J OR Wnet= Fnetd =(Fa-Ff)d =(100cos37-50)(40m) FN Ff Fa Fg
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FORCE VS DISPLACEMENT Work is the area under the curve. F” (N) d (m) A=length*width=F”d=W A=1/2*base *height =1/2dF” =-W A=1/2(p+q)h trapezoid
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