JEE Main & Advanced Physics Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति Question Bank Self Evaluation Test - Work, Energy and Power

  • question_answer
    A 2 kg block slides on a horizontal floor with a speed of4m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15N and spring constant is 10,000 N/m. The spring compresses by Critical Thinking

    A) 8.5 cm

    B) 5.5 cm

    C) 2.5 cm

    D) 11.0 cm

    Correct Answer: B

    Solution :

    [b] Let the blow compress the spring by x before stopping. Kinetic energy of the block = (P.E of compressed spring) + work done against function. \[\frac{1}{2}\times 2\times {{\left( 4 \right)}^{2}}=\frac{1}{2}\times 10,000\times {{x}^{2}}+\left( +15 \right)\times x\] \[10,000{{x}^{2}}+30x-32=0\] \[\Rightarrow 5,000{{x}^{2}}+15x-16=0\] \[\therefore \,\,\,x=-\frac{15\pm \sqrt{{{\left( 15 \right)}^{2}} \pm  - 4\times (5000)(-16)}}{2\times 5000}\] \[=0.055\text{ }m=5.5\text{ }cm.\]


You need to login to perform this action.
You will be redirected in 3 sec spinner