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专题三　计算题突破
INCLUDEPICTURE "课后检测能力达标.tif" INCLUDEPICTURE "课后检测能力达标.tif" \* MERGEFORMAT INCLUDEPICTURE "课后检测能力达标.tif" \* MERGEFORMAT INCLUDEPICTURE "课后检测能力达标.tif" \* MERGEFORMAT INCLUDEPICTURE "课后检测能力达标.tif" \* MERGEFORMAT INCLUDEPICTURE "课后检测能力达标.tif" \* MERGEFORMAT
1．(7分)如图所示，在某次冰壶运动中，运动员用30 N的水平力将冰壶从A点推至B点后放手，用时2 s，随后离手后的冰壶沿水平方向继续滑行至O点停止，用时8 s．如图乙所示，A点到O点的距离是30 m，A点到B点的距离是4 m．求：
(1)冰壶从A点到O点的平均速度；(3分)
(2)投掷过程中运动员对冰壶做功的功率．(4分)
INCLUDEPICTURE "25W-626.tif" INCLUDEPICTURE "25W-626.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-626.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-626.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-626.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-627.tif" INCLUDEPICTURE "25W-627.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-627.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-627.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-627.tif" \* MERGEFORMAT
甲 乙
2．(8分)重力为1 680 N的建筑材料，放在工地的水平地面上，用如图所示的滑轮组将建筑材料匀速提升了3 m，滑轮组的机械效率是84%.试求：
(1)滑轮组做的有用功是多少？(4分)
(2)电动机对钢丝绳的拉力F是多大？(4分)
INCLUDEPICTURE "25W-628.tif" INCLUDEPICTURE "25W-628.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-628.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-628.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-628.tif" \* MERGEFORMAT
3．(7分)近两年，新能源汽车发展迅速，如图所示，一辆新能源汽车搭载了能量为45 kW·h的电池包，在某次测试中，该车匀速行驶了40 km，用时30 min，电池的电量从80%下降到70%，匀速行驶时所受的阻力为350 N．求：
(1)该电动车牵引力做的功；(3分)
(2)该电动车工作的效率是多少？(小数点后保留一位小数)(4分)
INCLUDEPICTURE "25W-624.tif" INCLUDEPICTURE "25W-624.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-624.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-624.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-624.tif" \* MERGEFORMAT
4．(8分)如图甲所示，吊车利用伸缩撑杆可使吊臂绕O点缓慢转动向上起吊货物．起吊时，伸缩撑杆对吊臂的支持力F始终与吊臂垂直，且作用点不变．将其简化成如图乙所示的杠杆，不计吊臂和钢丝绳的自重，若货物总质量为1 t，l1＝0.5 m，l2＝1.6 m，货物上升的高度为50 cm，g取10 N/kg.求在此过程中：
(1)伸缩撑杆对吊臂的支持力F；(2分)
(2)钢丝绳对货物拉力所做的功；(3分)
(3)试分析说明在吊起货物的过程中，随着吊臂的升高，支持力F大小如何变化？(3分)
INCLUDEPICTURE "25W-629.tif" INCLUDEPICTURE "25W-629.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-629.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-629.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-629.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-630.tif" INCLUDEPICTURE "25W-630.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-630.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-630.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-630.tif" \* MERGEFORMAT
甲 乙
5．(8分)(2025·江西模拟预测)如图所示，“潜龙三号”是我国自主研制的潜水器，长3.5 m，高1.5 m，平均宽度约1 m，体积约6 m3，最大下潜深度为4 000 m，体形像“小丑鱼”．在首次海试中：(ρ海水取1.0×103 kg/m3，g取10 N/kg)
INCLUDEPICTURE "26JXWL-293.tif" INCLUDEPICTURE "26JXWL-293.tif" \* MERGEFORMAT INCLUDEPICTURE "26JXWL-293.tif" \* MERGEFORMAT INCLUDEPICTURE "26JXWL-293.tif" \* MERGEFORMAT INCLUDEPICTURE "26JXWL-293.tif" \* MERGEFORMAT
(1)求潜水器下潜至3 900 m深处时受到海水的压强；(2分)
(2)求潜水器浸没时受到海水的浮力；(3分)
(3)如果此潜水器下潜到4 000 m以下的海底，潜水器与水底紧密接触，请你分析将会造成什么后果．(3分)
6．(8分)如图甲所示，为家庭电路的一部分，两个灯泡的U－I图像如图丙所示，关闭家里其他所有用电器同时闭合开关S1和S2，两灯正常发光．求：
(1) 灯L1正常发光的电阻；(2分)
(2)流经电能表的电流；(3分)
(3)电路工作20 min，电能表的表盘转多少圈？(3分)
INCLUDEPICTURE "25W-634.tif" INCLUDEPICTURE "25W-634.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-634.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-634.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-634.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-635.tif" INCLUDEPICTURE "25W-635.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-635.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-635.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-635.tif" \* MERGEFORMAT
甲 乙
INCLUDEPICTURE "25W-636.tif" INCLUDEPICTURE "25W-636.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-636.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-636.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-636.tif" \* MERGEFORMAT
丙
7．(8分)一台家用电水壶(图甲)，有加热和保温两种功能．其内部电路的简图如图乙所示，R1、R2均为加热电阻，通过单刀双掷开关可以实现加热和保温两种功能的切换．电水壶加热功率为1 000 W，保温功率为44 W.
(1)单刀双掷开关旋转到1时是加热状态还是保温状态，请分析说明；(3分)
(2)若电流产生的热量全部被水吸收，把500 g的水从40 ℃加热到100 ℃，使用电水壶的加热挡需要的时间[水的比热容c水＝4.2×103J/(kg·℃)]；(2分)
(3)求R1和R2的阻值为多少．(3分)
INCLUDEPICTURE "25W-637.tif" INCLUDEPICTURE "25W-637.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-637.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-637.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-637.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-638.tif" INCLUDEPICTURE "25W-638.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-638.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-638.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-638.tif" \* MERGEFORMAT
甲 乙
8．(8分)坐位体前屈是某市体育中考的抽考项目之一．坐位体前屈项目测试仪原理如图所示，电源电压恒为10 V，R1＝50 Ω，电阻丝R2粗细均匀，总长度为30 cm，R2的阻值和长度成正比．闭合开关，挡板P在B端时，坐位体前屈成绩为0 cm，电流表示数为0.05 A；挡板P在A端时，坐位体前屈成绩为30 cm，求：
(1)电阻丝R2的最大阻值．(2分)
(2)挡板P在A端时电压表的示数．(3分)
(3)规定男生坐位体前屈成绩21.6 cm以上为优秀，若小明进行测试时电压表示数为4 V，通过计算判断小明的成绩是否为优秀？(3分)
INCLUDEPICTURE "25W-632.tif" INCLUDEPICTURE "25W-632.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-632.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-632.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-632.tif" \* MERGEFORMAT专题三　计算题突破
INCLUDEPICTURE "课后检测能力达标.tif" INCLUDEPICTURE "课后检测能力达标.tif" \* MERGEFORMAT INCLUDEPICTURE "课后检测能力达标.tif" \* MERGEFORMAT INCLUDEPICTURE "课后检测能力达标.tif" \* MERGEFORMAT INCLUDEPICTURE "课后检测能力达标.tif" \* MERGEFORMAT INCLUDEPICTURE "课后检测能力达标.tif" \* MERGEFORMAT
1．(7分)如图所示，在某次冰壶运动中，运动员用30 N的水平力将冰壶从A点推至B点后放手，用时2 s，随后离手后的冰壶沿水平方向继续滑行至O点停止，用时8 s．如图乙所示，A点到O点的距离是30 m，A点到B点的距离是4 m．求：
(1)冰壶从A点到O点的平均速度；(3分)
(2)投掷过程中运动员对冰壶做功的功率．(4分)
INCLUDEPICTURE "25W-626.tif" INCLUDEPICTURE "25W-626.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-626.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-626.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-626.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-627.tif" INCLUDEPICTURE "25W-627.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-627.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-627.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-627.tif" \* MERGEFORMAT
甲 乙
解：(1)由题意可得，冰壶从A点到O点所用的时间为
t＝2 s＋8 s＝10 s
由v＝可得，冰壶从A点到O点的平均速度为v＝＝3 m/s.
(2)由题意可得，AB段的距离为冰壶受到推力移动的距离，则这次投掷过程中运动员对冰壶做的功为
W＝Fs＝30 N×4 m＝120 J
这次投掷过程中运动员对冰壶做功的功率为
P＝＝＝60 W.
2．(8分)重力为1 680 N的建筑材料，放在工地的水平地面上，用如图所示的滑轮组将建筑材料匀速提升了3 m，滑轮组的机械效率是84%.试求：
(1)滑轮组做的有用功是多少？(4分)
(2)电动机对钢丝绳的拉力F是多大？(4分)
INCLUDEPICTURE "25W-628.tif" INCLUDEPICTURE "25W-628.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-628.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-628.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-628.tif" \* MERGEFORMAT
解：(1)滑轮组做的有用功
W有＝Gh＝1 680 N×3 m＝5 040 J.
(2)由图可知，钢丝绳的有效股数n＝2，钢丝绳端通过的距离
s＝nh＝2×3 m＝6 m
由η＝得到总功为
W总＝＝＝6 000 J
故由W＝Fs得电动机对钢丝绳的拉力F为
F＝＝＝1 000 N.
3．(7分)近两年，新能源汽车发展迅速，如图所示，一辆新能源汽车搭载了能量为45 kW·h的电池包，在某次测试中，该车匀速行驶了40 km，用时30 min，电池的电量从80%下降到70%，匀速行驶时所受的阻力为350 N．求：
(1)该电动车牵引力做的功；(3分)
(2)该电动车工作的效率是多少？(小数点后保留一位小数)(4分)
INCLUDEPICTURE "25W-624.tif" INCLUDEPICTURE "25W-624.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-624.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-624.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-624.tif" \* MERGEFORMAT
解：(1)电动车匀速运动，受力平衡，因此牵引力等于阻力，即F＝f＝350 N
牵引力做的功
W＝Fs＝350 N×4×104m＝1.4×107J.
(2)由题意知，消耗的电能
W电＝(80%－70%)×45 kW·h＝4.5 kW·h＝1.62×107J
电动车工作的效率
η＝×100%＝×100%≈86.4%.
4．(8分)如图甲所示，吊车利用伸缩撑杆可使吊臂绕O点缓慢转动向上起吊货物．起吊时，伸缩撑杆对吊臂的支持力F始终与吊臂垂直，且作用点不变．将其简化成如图乙所示的杠杆，不计吊臂和钢丝绳的自重，若货物总质量为1 t，l1＝0.5 m，l2＝1.6 m，货物上升的高度为50 cm，g取10 N/kg.求在此过程中：
(1)伸缩撑杆对吊臂的支持力F；(2分)
(2)钢丝绳对货物拉力所做的功；(3分)
(3)试分析说明在吊起货物的过程中，随着吊臂的升高，支持力F大小如何变化？(3分)
INCLUDEPICTURE "25W-629.tif" INCLUDEPICTURE "25W-629.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-629.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-629.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-629.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-630.tif" INCLUDEPICTURE "25W-630.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-630.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-630.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-630.tif" \* MERGEFORMAT
甲 乙
解：(1)货物的重力为
G＝mg＝1×103kg×10 N/kg＝1×104N
由杠杠平衡条件Fl1＝Gl2可知，伸缩撑杆对吊臂的支持力
F＝＝＝3.2×104N.
(2)由题意可知，钢丝绳对货物拉力所做的功等于克服货物重力做功，则
W＝Fs＝Gh＝1×104N×0.5 m＝5×103J.
(3)重力的方向竖直向下，在吊起货物的过程中，随着吊臂的升高，货物的重力不变，所以吊臂受到的阻力不变，阻力臂变小，动力臂也不变，由杠杠平衡条件F1l1＝F2l2可知，支持力F变小．
5．(8分)(2025·江西模拟预测)如图所示，“潜龙三号”是我国自主研制的潜水器，长3.5 m，高1.5 m，平均宽度约1 m，体积约6 m3，最大下潜深度为4 000 m，体形像“小丑鱼”．在首次海试中：(ρ海水取1.0×103 kg/m3，g取10 N/kg)
INCLUDEPICTURE "26JXWL-293.tif" INCLUDEPICTURE "26JXWL-293.tif" \* MERGEFORMAT INCLUDEPICTURE "26JXWL-293.tif" \* MERGEFORMAT INCLUDEPICTURE "26JXWL-293.tif" \* MERGEFORMAT INCLUDEPICTURE "26JXWL-293.tif" \* MERGEFORMAT
(1)求潜水器下潜至3 900 m深处时受到海水的压强；(2分)
(2)求潜水器浸没时受到海水的浮力；(3分)
(3)如果此潜水器下潜到4 000 m以下的海底，潜水器与水底紧密接触，请你分析将会造成什么后果．(3分)
解：(1)潜水器下潜至3 900 m深处时受到海水的压强
p＝ρ海水gh＝1.0×103 kg/m3×10 N/kg×3 900 m＝3.9×107 Pa.
(2)潜水器浸没时受到海水的浮力
F浮＝G排＝m排g＝ρ海水gV排＝1.0×103 kg/m3×10 N/kg×6 m3＝6×104 N.
(3)根据p＝ρgh可知如果此潜水器下潜到4 000 m以下的海底，则可能超过其最大压强承受范围，潜水艇受到极大的向下压力，且潜水器与水底紧密接触，由浮力产生原因可知F浮＝F向上－F向下，此时F向上为零，F向下极大，则潜水艇可能无法自主上升．
6．(8分)如图甲所示，为家庭电路的一部分，两个灯泡的U－I图像如图丙所示，关闭家里其他所有用电器同时闭合开关S1和S2，两灯正常发光．求：
(1) 灯L1正常发光的电阻；(2分)
(2)流经电能表的电流；(3分)
(3)电路工作20 min，电能表的表盘转多少圈？(3分)
INCLUDEPICTURE "25W-634.tif" INCLUDEPICTURE "25W-634.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-634.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-634.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-634.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-635.tif" INCLUDEPICTURE "25W-635.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-635.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-635.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-635.tif" \* MERGEFORMAT
甲 乙
INCLUDEPICTURE "25W-636.tif" INCLUDEPICTURE "25W-636.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-636.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-636.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-636.tif" \* MERGEFORMAT
丙
解：(1)由丙图可知，当L1两端的电压U＝220 V，通过它的电流I1＝0.25 A，故灯L1正常发光的电阻为
R1＝＝＝880 Ω.
(2)电能表在干路，通过电能表的电流等于通过L1、L2的电流之和．由丙图可知，两灯正常工作时，通过L1、L2的电流分别是I1＝0.25 A，I2＝0.4 A，故通过电能表的电流为
I＝I1＋I2＝0.25 A＋0.4 A＝0.65 A．
(3)L1、L2的功率为
P1＝UI1＝220×0.25 A＝55 W＝0.055 kW，P2＝UI2＝220×0.4 A＝88 W＝0.088 kW
电路工作20 min，消耗的电能为
W＝(P1＋P2)t＝(0.055 kW＋0.088 kW)× h＝ kW·h
电能表的表盘转过的圈数为
n＝WN＝×3 000 r＝143 r.
7．(8分)一台家用电水壶(图甲)，有加热和保温两种功能．其内部电路的简图如图乙所示，R1、R2均为加热电阻，通过单刀双掷开关可以实现加热和保温两种功能的切换．电水壶加热功率为1 000 W，保温功率为44 W.
(1)单刀双掷开关旋转到1时是加热状态还是保温状态，请分析说明；(3分)
(2)若电流产生的热量全部被水吸收，把500 g的水从40 ℃加热到100 ℃，使用电水壶的加热挡需要的时间[水的比热容c水＝4.2×103J/(kg·℃)]；(2分)
(3)求R1和R2的阻值为多少．(3分)
INCLUDEPICTURE "25W-637.tif" INCLUDEPICTURE "25W-637.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-637.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-637.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-637.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-638.tif" INCLUDEPICTURE "25W-638.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-638.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-638.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-638.tif" \* MERGEFORMAT
甲 乙
解：(1)分析电路可知，闭合开关S，单刀双掷开关旋转到1时，R1、R2串联，此时电路总电阻较大，根据P＝可知总功率较小，为保温挡；闭合开关S，单刀双掷开关旋转到2时，只有R2工作，此时电路总电阻较小，根据P＝可知总功率较大，为加热挡．
(2)水吸收的热量为
Q吸＝c水mΔt＝4.2×103J/(kg·℃)×0.5 kg×(100 ℃－40 ℃)＝1.26×105J
电流产生的热量全部被水吸收，则需要的时间为
t＝＝＝＝126 s.
(3)加热挡时，只有R2工作，则R2的电阻为
R2＝＝＝48.4 Ω
保温挡时，R1、R2串联，总电阻为
R总＝＝＝1 100 Ω
则电阻R1的阻值为
R1＝R总－R2＝1 100 Ω－48.4 Ω＝1 051. 6 Ω.
8．(8分)坐位体前屈是某市体育中考的抽考项目之一．坐位体前屈项目测试仪原理如图所示，电源电压恒为10 V，R1＝50 Ω，电阻丝R2粗细均匀，总长度为30 cm，R2的阻值和长度成正比．闭合开关，挡板P在B端时，坐位体前屈成绩为0 cm，电流表示数为0.05 A；挡板P在A端时，坐位体前屈成绩为30 cm，求：
(1)电阻丝R2的最大阻值．(2分)
(2)挡板P在A端时电压表的示数．(3分)
(3)规定男生坐位体前屈成绩21.6 cm以上为优秀，若小明进行测试时电压表示数为4 V，通过计算判断小明的成绩是否为优秀？(3分)
INCLUDEPICTURE "25W-632.tif" INCLUDEPICTURE "25W-632.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-632.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-632.tif" \* MERGEFORMAT INCLUDEPICTURE "25W-632.tif" \* MERGEFORMAT
解：(1)闭合开关，挡板P在B端时，坐位体前屈成绩为0 cm，电流表示数为0.05 A，R1两端的电压
U1＝IR1＝0.05 A×50 Ω＝2.5 V
电阻丝R2两端的电压
U2＝U－U1＝10 V－2.5 V＝7.5 V
电阻丝R2的最大阻值
R2＝＝＝150 Ω.
(2)由图可以看出，挡板P在A端时电压表测R2两端的电压，此时电压表的示数为7.5 V.
(3)小明进行测试时，电阻丝R2接入电路中的电阻
R2′＝＝＝80 Ω
因电阻丝R2的阻值和长度成正比，所以，接入电路中的电阻丝的长度：
×30 cm＝16 cm
即小明坐位体前屈的成绩为16 cm，小于21.6 cm，所以，小明的成绩不是优秀．

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