A body of mass 0.25 kg travels along a straight line from \( x = 0 \) to \( x = 2 \, \text{m} \) with a speed \( v = k x^2 \) where \( k = 2 \, \text{m}^{-1} \). The work done by the net force during this displacement is
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When calculating the work done by a force, the change in kinetic energy is often the easiest method to apply, especially when velocity is given as a function of position.
The speed \( v \) is given by the equation \( v = k x^2 \), where \( k \) is a constant. To find the work done by the net force, we use the work-energy theorem, which states that the work done by the net force is equal to the change in kinetic energy.
The kinetic energy \( K \) of the body at position \( x \) is given by:
\[
K = \frac{1}{2} m v^2
\]
Substituting the expression for \( v \):
\[
K = \frac{1}{2} m (k x^2)^2 = \frac{1}{2} m k^2 x^4
\]
Now, the work done by the net force is the change in kinetic energy as the body moves from \( x = 0 \) to \( x = 2 \, \text{m} \). The change in kinetic energy is given by:
\[
W = K_2 - K_1 = \frac{1}{2} m k^2 x_2^4 - 0
\]
Substitute \( m = 0.25 \, \text{kg} \), \( k = 2 \, \text{m}^{-1} \), and \( x_2 = 2 \, \text{m} \):
\[
W = \frac{1}{2} (0.25) (2^2)^4 = \frac{1}{2} (0.25) \times 16 = 4 \, \text{J}
\]
Thus, the work done by the net force is \( 4 \, \text{J} \).
