{"id":326989,"date":"2023-01-05T23:56:17","date_gmt":"2023-01-05T18:26:17","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/question\/a-block-of-mass-10-kg-moving-in-x-direction-with-a-constant-speed-of-10-ms-1-is-subjected-to-a-retarding-force-f-0-1x-j-m-during-its-travel-from-x-20-m\/"},"modified":"2025-06-26T17:53:30","modified_gmt":"2025-06-26T12:23:30","slug":"a-block-of-mass-10-kg-moving-in-x-direction-with-a-constant-speed-of-10-ms-1-is-subjected-to-a-retarding-force-f-0-1x-j-m-during-its-travel-from-x-20-m","status":"publish","type":"questions","link":"https:\/\/infinitylearn.com\/surge\/question\/physics\/a-block-of-mass-10-kg--moving-in-x-direction-with-a-cons\/","title":{"rendered":"A block of mass \u00a010 kg , moving in \u00a0x \u00a0direction with a constant \u00a0speed of 10\u00a0ms-1, is subjected to a retarding force \u00a0F = 0.1x J\/m during its travel from \u00a0x = 20 m to \u00a030 m. Its final KE will be\u00a0"},"content":{"rendered":"","protected":false},"author":1,"template":"","meta":{"_yoast_wpseo_focuskw":"A block of mass 10 kg moving in \u00a0x \u00a0direction","_yoast_wpseo_title":"","_yoast_wpseo_metadesc":"Here is the answer C - 475 J. Click here to know the complete solution, please visit us at infinitylearn.com","custom_permalink":"question\/physics\/a-block-of-mass-10-kg--moving-in-x-direction-with-a-cons\/"},"categories":[],"acf":{"question":"

A block of mass  10 kg , moving in  x  direction with a constant  speed of 10 m<\/mi>s<\/mi>-<\/mo>1<\/mn><\/mrow><\/msup><\/math>, is subjected to a retarding force  F = 0.1x J\/m during its travel from  x = 20 m to  30 m. Its final KE will be <\/p>","options":[{"option":"

275 J<\/p>","correct":false},{"option":"

250 J<\/p>","correct":false},{"option":"

475 J<\/p>","correct":true},{"option":"

450 J<\/p>","correct":false}],"solution":"

 Here, <\/mtext>m<\/mi>=<\/mo>10<\/mn> <\/mo>kg<\/mi>,<\/mo>v<\/mi>i<\/mi><\/msub>=<\/mo>10<\/mn> <\/mo>m<\/mi> <\/mo>s<\/mi><\/mrow>-<\/mo>1<\/mn><\/mrow><\/msup><\/math><\/p>

Initial kinetic energy of the block is <\/p>

K<\/mi>i<\/mi><\/msub>=<\/mo>1<\/mn>2<\/mn><\/mfrac>m<\/mi>v<\/mi>i<\/mi>2<\/mn><\/msubsup>=<\/mo>1<\/mn>2<\/mn><\/mfrac>\u00d7<\/mo>(<\/mo>10<\/mn> <\/mo>kg<\/mi>)<\/mo>\u00d7<\/mo>10<\/mn> <\/mo>m<\/mi> <\/mo>s<\/mi><\/mrow>-<\/mo>1<\/mn><\/mrow><\/msup><\/mrow><\/mfenced>2<\/mn><\/msup>=<\/mo>500<\/mn> <\/mo>J<\/mi><\/math><\/p>

work done by retarding force<\/p>

W<\/mi>=<\/mo>\u222b<\/mo>x<\/mi>1<\/mn><\/msub>x<\/mi>2<\/mn><\/msub><\/msubsup>F<\/mi>r<\/mi><\/msub>d<\/mi>x<\/mi>=<\/mo>\u222b<\/mo>20<\/mn>30<\/mn><\/msubsup>-<\/mo>0<\/mn>.<\/mo>1<\/mn>x<\/mi>d<\/mi>x<\/mi>=<\/mo>-<\/mo>0<\/mn>.<\/mo>1<\/mn>x<\/mi>2<\/mn><\/msup>2<\/mn><\/mfrac><\/mfenced>20<\/mn>30<\/mn><\/msubsup><\/math><\/p>

=<\/mo>-<\/mo>0<\/mn>.<\/mo>1<\/mn>900<\/mn>-<\/mo>400<\/mn><\/mrow>2<\/mn><\/mfrac><\/mfenced>=<\/mo>-<\/mo>25<\/mn> <\/mo>J<\/mi><\/math><\/p>

According to work energy theorem,<\/p>

W<\/mi>=<\/mo>K<\/mi>f<\/mi><\/msub>-<\/mo>K<\/mi>i<\/mi><\/msub><\/mtd><\/mtr>K<\/mi>f<\/mi><\/msub>=<\/mo>W<\/mi>+<\/mo>K<\/mi>i<\/mi><\/msub>=<\/mo>-<\/mo>25<\/mn> <\/mo>J<\/mi>+<\/mo>500<\/mn> <\/mo>J<\/mi>=<\/mo>475<\/mn> <\/mo>J<\/mi><\/mtd><\/mtr><\/mtable><\/math><\/p>","subject":""},"yoast_head":"\nA block of mass \u00a010 kg , moving in \u00a0x \u00a0direction with a constant \u00a0speed of 10\u00a0ms-1, is subjected to a retarding force \u00a0F = 0.1x J\/m during its travel from \u00a0x = 20 m to \u00a030 m. Its final KE will be\u00a0 - Infinity Learn by Sri Chaitanya<\/title>\n<meta name=\"description\" content=\"Here is the answer C - 475 J. 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