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The value of the acting force in the given diagram isAcceleration = 0.1 ms-2Mass = 10 kg |
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Answer» For 10kg block, f is frictional forcef 1max =μ 1 μ 2 =μ 1 (10G)=0.1(10×10)=10N i) if F=2N then acting friction f 1 =F=2N(A) is correct.for 5 KG block;f 2max =μ(10+5)g=0.3×15×10=45Nii) if F=30N then; F−F 1max =10a10a=30−10=20a=2m/s 2 (B) is correct.iii) if μ 1 changed to 0.5;then f 1max =μ 1 (10g)=0.5×10×10=50NIn the case acting friction will be F 2max Since,F 1max =F 2max Block 10kg and 5kg both move together then F min to begin isF min −F max =0 F min =F 2max =45N (c) is correct.iv) if μ=0.1 , then f max =10N and,f 2max =45Nf 2max >f 1max always for any any value of N. Hence, 5kg block never move on the ground.So, OPTION (A), (B), (C), and (D) are correct.xplanation: |
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