Question:
A block of mass 1 kg moving with velocity varies according to position as \( v = 2x^2 \). If the block goes from \( x = 0 \) to \( x = 5 \), find the work done by the block.
A block of mass 1 kg moving with velocity varies according to position as \( v = 2x^2 \). If the block goes from \( x = 0 \) to \( x = 5 \), find the work done by the block.
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When the velocity is given as a function of position, the work done can be calculated using integration. Work can also be found using the change in kinetic energy, \( W = \Delta K.E \).
Updated On: Apr 7, 2026
- zero
- 1250 J
- 1000 J
- 750 J
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The Correct Option is B
Solution and Explanation
The work done \( W \) is given by the equation:
\[
W = \Delta KE = \frac{1}{2} M \left[ V_2^2 - V_1^2 \right]
\]
At \( x = 0, V_1 = 0 \), and at \( x = 5, V_2 = 2 \times (5)^2 = 50 \).
Now, substitute these values into the equation for \( W \):
\[
W = \frac{1}{2} \times 1 \times 1 \left[ 50^2 - 0^2 \right]
\]
\[
W = \frac{1}{2} \times 2500 = 1250 \, \text{T}
\]
Thus, the work done is \( W = 1250 \, \text{T} \).
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