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AIIMS 2014
Exam
AIIMS
Year
2014
Subject
Physics
Chapter
Work, Power and Energy

AIIMS 2014 · Physics · Work, Power and Energy

A spring of spring constant 5 × 10^3 ~N ~m^ -1 is stretched...

A spring of spring constant 5×103 N m−15 \times 10^3 \mathrm{~N} \mathrm{~m}^{-1} is stretched initially by 5 cm5 \mathrm{~cm} from the unstretched position. The work required to stretch it further by another 5 cm5 \mathrm{~cm} is

Options

  1. A6.25 N m6.25 \mathrm{~N} \mathrm{~m}
  2. B1250 N m1250 \mathrm{~N} \mathrm{~m}
  3. C18.75 N m18.75 \mathrm{~N} \mathrm{~m}
  4. D25.00 N m25.00 \mathrm{~N} \mathrm{~m}

Correct answer

C. 18.75 N m18.75 \mathrm{~N} \mathrm{~m}

Step-by-step solution

U1=12kx12=12×U_1=\frac{1}{2} k x_1^2=\frac{1}{2} \times\left(5 \times 10^3\right) \times(5×10−2)2\left(5 \times 10^{-2}\right)^2 =6.25 N m=6.25 \mathrm{~N} \mathrm{~m} U2=12kx22=12×5×103×(5+5)2×10−4U_2=\frac{1}{2} k x_2^2=\frac{1}{2} \times 5 \times 10^3 \times(5+5)^2 \times 10^{-4} =25 N m=25 \mathrm{~N} \mathrm{~m} ∴\therefore Work done =U2−U1=25.0−6.25=U_2-U_1=25.0-6.25 =18.75 N m=18.75 \mathrm{~N} \mathrm{~m}
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