Question

某同學(xué)用位移傳感器研究木塊在斜面上的滑動(dòng)情況,裝置如圖(a).已知斜面傾角θ=37°.他使木塊以初速度v₀沿斜面上滑,并同時(shí)開(kāi)始記錄數(shù)據(jù),繪得木塊從開(kāi)始上滑至最高點(diǎn),然后又下滑回到出發(fā)處過(guò)程中的x -t圖線如圖(b)所示.圖中曲線左側(cè)起始端的坐標(biāo)為(0,1.4),曲線最低點(diǎn)的坐標(biāo)為(0.5,0.4).重力加速度g取10 m/s².sin37°=0.6,cos37°=0.8.求:

 (1)木塊上滑時(shí)的初速度v₀和上滑過(guò)程中的加速度a₁;

(2)木塊與斜面間的動(dòng)摩擦因數(shù)μ;

(3)木塊滑回出發(fā)點(diǎn)時(shí)的速度v.

Answer 1

【解析】(1)0～0.5 s段曲線滿足拋物線方程,由圖線得

x=k(t-0.5)²+0.4                                               (1分)

將(0,1.4)代入得1.4=k(0-0.5) ²+0.4,解得k=4                     (1分)

x=4(t-0.5)²+0.4=4t²-4t+1.4                                  (1分)

所以v₀=4 m/s                                               (1分)

a₁=8 m/s²                                                   (1分)

(2)上滑過(guò)程

mgsin37°+μmgcos37°=ma                                   (2分)

解得μ=0.25                                                (1分)

(3)下滑過(guò)程

mgsin37°-μmgcos37°=ma₂                                   (2分)

a₂=4 m/s²                                                   (1分)

由v²=2a₂x得 m/s                              (1分)

答案:(1)4 m/s    8 m/s²      (2)0.25    (3)2.83 m/s