Q: (1) power P=? ; (2) The mechanical efficiency of pulley block η =? ; (3) The minimum pressure of an object on the ground p=?
Solution: (1) Tension moving speed: V pull =2v object =2×0.2m/s=0.4m/s,
The power of the worker's master to pull the rope:
p = Wt = Fst = Fv = 600n×0.4m/s = 240 w；
(2) Mechanical efficiency:
η=W Useful w total ×100% = ghf2h×100% = g2f×100% =140n2× 600n×100%.
(3) excluding friction and rope weight, F= 12(G+G motion),
So the moving pulley is heavy:
g move = 2f-g = 2×600n- 1 140n = 60n，
The maximum pulling force when Xiao Ming pulls the rope is equal to its gravity, f ′ = g = 450n, f ′ = g12 (f pulling +G moving).
So the maximum pulling force of the rope on the object: F pulling force =2F'-G moving =2×450N-60N=840N.
Minimum pressure of objects on the ground:
F pressure =F branch =G-F tension = 1 140N-840N=300N,
Minimum pressure of objects on the ground:
P=F pressure s = 300n0. 15m2 = 2000pa.
Answer: (1) The power of the master pulling the rope is 240 W；;
(2) The mechanical efficiency of pulley block is 95%;
(3) The minimum pressure of objects on the ground is 2000 Pa. ..