s′= 2s = 2×0.25m = 0.5m，
The work done by a person to pull the drill bow back and forth to overcome friction;
w = Fs′= 16n×0.5m = 8J；
(2) From the topic, Q absorption = w× 25% = 8 j× 25% = 2 j;
∫Q absorption =cm△t, m=0.25g=0.25× 10-3kg,
The temperature of the walking stick has risen;
△t=Q absorption cm=2J2× 103J/(kg? ℃)×0.25× 10? 3kg？ =4℃;
(3) As can be seen from (2), the temperature of the stick increases by 4℃ per second,
Time required to raise the temperature from 260℃ to 20℃ = 240℃;
t = 240℃±4℃= 60s。
Answer: (1) The work done by a person to pull the drill bow back and forth to overcome friction is 8j;
(2) In 65438 0 seconds, the tip temperature of the rod can be increased by 4℃;
(3) It takes about 60s to make the stick burn.