Question

如圖1-3-10所示，質(zhì)量為10 kg的物體在F=200 N的水平推力作用下，從粗糙斜面的底端由靜止開始沿斜面運(yùn)動(dòng)，斜面固定不動(dòng)，與水平面的夾角θ=37°.力F作用2秒鐘后撤去，物體在斜面上繼續(xù)上滑了1.25秒后，速度減為零.求：物體與斜面間的動(dòng)摩擦因數(shù)μ和物體的總位移s. （已知sin37°=0.6，cos37°=0.8，g=10 m/s²）

圖1-3-10

 

Answer 1

解析：力F作用時(shí)有Fcosθ-μF_N-mgsinθ=ma                                   ①

Fsinθ+mgcosθ=F_N                                                                                        ②

2 s時(shí)物體的速度為v，v=2a                                                   ③

位移為s₁，  s₁=2a，                                                         ④

撤去力F后，有mgsinθ+μmgcosθ=ma′，                                    ⑤

v=1.25a′                                                                   ⑥

位移為s₂，s₂=a′                                                     ⑦

由③⑥得a=0.625a′，聯(lián)立①②⑤得，μ=0.25.代入①⑤得a=5 m/s²，a′=8 m/s².由④⑦得s₁=10 m，s₂=6.25 m

總位移s=s₁+s₂=16.25 m

答案：μ=0.25，s=16.25 m