Question

If the work done to double the velocity of a body from \(15\ \mathrm{ms^{-1}}\) is \(K\) times the work done to double its velocity from \(10\ \mathrm{ms^{-1}}\), then the value of \(K\) is

• A. \(2.75\)
• B. \(2.25\)
• C. \(3.25\)
• D. \(3.75\)

Hint:

If a body's speed is doubled, \[ \boxed{ W=\Delta KE =\frac12m\left((2u)^2-u^2\right) =\frac32mu^2. } \] Thus, the work required is directly proportional to the square of the initial speed.

Answer 1

The Correct Option is B

Step 1: Use the work-energy theorem. Work done is equal to the change in kinetic energy: \[ W=\frac12m(v^2-u^2). \] If the velocity is doubled from \[ u, \] then \[ v=2u. \] Hence, \[ W = \frac12m\left((2u)^2-u^2\right) = \frac32mu^2. \]

Step 2: Find the work done in each case. For \[ u=15\ \mathrm{ms^{-1}}, \] \[ W_1 = \frac32m(15)^2. \] For \[ u=10\ \mathrm{ms^{-1}}, \] \[ W_2 = \frac32m(10)^2. \] Therefore, \[ K = \frac{W_1}{W_2} = \frac{15^2}{10^2} = \frac{225}{100} = 2.25. \]

Step 3: Write the answer. Hence, \[ \boxed{2.25}. \] Thus, \[ \boxed{(B)} \] is the correct answer.