地面上有质量m=2.0×103 kg的物体，用起重机将它竖直向上匀加速提升到h="10" m的高处，物体在这段上升过程中，起重机钢绳对它的拉力F=2.1×104 N。取g="10" m/s2，将物体视为质点．
（1）求钢绳的拉力F所做的功WF。
（2）选择地面为参考平面，求物体在h="10" m处的重力势能EP。
（1）2.0×105J（2）2.0×105J
题目分析：（1）钢绳的拉力F所做的功WF＝Fh＝2. 1×105 J（4分）
（2）物体的重力势能EP＝mgh＝2.0×105J（3分）
说明：第（1）问中写出公式记3分，得出答案记1分；第（2）问中写出公式记2分，得出答案记1分．
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