Question

A car accelerates from rest to \( u \, \text{m/s} \). The energy spent in this process is \( E \, \text{J} \). The energy required to accelerate the car from \( u \, \text{m/s} \) to \( 2u \, \text{m/s} \) is \( nE \, \text{J} \). The value of \( n \) is.

Answer 1

Correct Answer: 3

To find the value of \( n \), we start by considering the kinetic energy formula:

\( KE = \frac{1}{2}mv^2 \)

where \( m \) is the mass and \( v \) is the velocity. Initially, the car accelerates from rest to velocity \( u \, \text{m/s} \). The kinetic energy at this velocity is:

\( KE_1 = \frac{1}{2}mu^2 = E \)

Next, the car accelerates from velocity \( u \) to \( 2u \). The kinetic energy at \( 2u \, \text{m/s} \) is:

\( KE_2 = \frac{1}{2}m(2u)^2 = \frac{1}{2}m(4u^2) = 2mu^2 \)

The energy required to increase the velocity from \( u \) to \( 2u \) is the difference in kinetic energy at these speeds:

\( \Delta KE = KE_2 - KE_1 = 2mu^2 - \frac{1}{2}mu^2 \)

\( \Delta KE = \frac{3}{2}mu^2 \)

Since the initial energy spent is \( E = \frac{1}{2}mu^2 \), we express the energy needed as a multiple of \( E \):

\( \Delta KE = 3 \times \frac{1}{2}mu^2 = 3E \)

Thus, the value of \( n \) is 3.

This result falls within the given range of 3,3, thus verifying the accuracy of our solution.