解:(1)∵F=0.3N,S=2m ∴W=FS=0.3N×2m=0.6J(2 )拉力所做的功为总功W总=W=0.6J∵G=1N,h=0.5m ∴W有用=Gh=1N×0.5m=0.5J从而可知,η= = ≈83.3%(3)∵秒表的指针转动了5个格,做功所需要的时间为:5×1s=5s ∴P= = =0.12W.
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