∴p=F Total S=2× 103N5× 10? 2 m2 = 4× 104 pa; ?
(2)∵g 1 = m 1g = 140kg× 10N/kg = 1.4× 103n,
∴w=g 1h= 1.4× 103n× 1.7m=2.38× 103j;
(3)P = Wt = 2.38× 103j 0.7s = 3.4× 103 w。 ?
Answer: (1) His pressure is 4×104 Pa;
(2) The work done by the athletes in lifting barbells this time is at least 2.38×103j; j;
(3) The power for athletes to lift barbells should be at least 3.4× 103W. w .