16 public questions tagged with this topic.
Work W=F·d=Fd cosθ per NCERT Chapter 6. Positive if force along displacement. Using given F, d, θ, calculation gives 54 J. Work-energy theorem W=ΔK confirms. So option B is correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Energy Conservation and Friction
Work W=F·d=Fd cosθ per NCERT Chapter 6. Positive if force along displacement. Using given F, d, θ, calculation gives 10 J. Work-energy theorem W=ΔK confirms. So option A is correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Energy Conservation and Friction
Work W=F·d=Fd cosθ per NCERT Chapter 6. Positive if force along displacement. Using given F, d, θ, calculation gives 700 J. Work-energy theorem W=ΔK confirms. So option B is correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Energy Conservation and Friction
Work W=F·d=Fd cosθ per NCERT Chapter 6. Positive if force along displacement. Using given F, d, θ, calculation gives 8 J. Work-energy theorem W=ΔK confirms. So option B is correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Laws of Motion and Energy Conservation
Work W=F·d=Fd cosθ per NCERT Chapter 6. Positive if force along displacement. Using given F, d, θ, calculation gives 3 J. Work-energy theorem W=ΔK confirms. So option A is correct.
Ref: NCERT Class 11 Physics > Chapter 6: Work, Energy and Power > Topic: Laws of Motion and Energy Conservation
Initial K=12×2×52=25J. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 25 J as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0
Initial K=12×3×82=96J. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 96 J as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0
Initial K=12×8×142=784J. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 784 J as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0
Initial K=12×2.5×72=61.25J. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 61.2 J as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0
Initial K=12×7×122=504J. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 504 J as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0
Initial K=12×4×92=162J. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 162 J as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0
Initial K=12×1.5×62=27J. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 27 J as the result, so option A is correct.
Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Practice Set0
