Question

Given: A particle with a mass of 0.5 kg moves along the x-axis and is braked by a horizontal braking force that provides the following braking acceleration: A_x (v_x) = -0.005 v_x² (m / s²)

The initial conditions at time t = 2 (s) are: x (2) = 25 (m) and v_x (2) = 40 (m / s)

The question that is asked is to calculate the amount of labour that the braking force performs between t = 2 (s) and t = 20 (s)

Answer 1

The initial kinetic energy at time t=2 s is

$KE_{initial}=\frac{1}{2} m V_{20}^2 = 0.5\times 0.5 kg \times (40 m/s)^2= 10J$

We are given with the acceleration as the function of velocity as :

$A_x(V_x)= -0.005 v_x^2 m/s^2$

(note: here since A is itself given in its proper units, that is, m/s^2, we infer that v_x is unitless, or only the magnitude of the velocity is considered here)

but

$A_x= \frac{\mathrm{d} v_x}{\mathrm{d} t}$

hence

$\frac{\mathrm{d} v_x}{\mathrm{d} t}= -0.005 v_x^2 m/s^2$

integrating this equation

$\int_{v_2}^{v_{20}} \frac{\mathrm{d} v_x }{v_x^2} = \int_{2}^{20} -0.005 \ \mathrm{d} t$

(note: dv_x has the unit m/s , however, v_x is having no unit as stated in the 'note' above.)

$\\ \int_{v_2}^{v_{20}} \frac{\mathrm{d} v_x }{v_x^2} m/s = \int_{t_i}^{t_f} -0.005 m/s^2\ \mathrm{d} t \\ \\ -\left [\frac{1}{v_{20}}- \frac{1}{v_2} \right ] m/s= -0.005 m/s^2 \times [t_f-t_i] \\ \\ \left [\frac{1}{v_{20}}- \frac{1}{40} \right ] m/s= 0.005 m/s^2\times [20 s-2s] \\ \\ \frac{1}{v_{20}}=\frac{1}{40}+0.005 \times 18= \frac{23}{200} \\ \\ \\v_{20}= \frac{200}{23}$

Hence velocity at t=20 s is $v_{20}=8.696 m/s$

hence final kinetic energy is

  $KE_{final}=\frac{1}{2} m V_{20}^2 = 0.5\times 0.5 kg \times (8.696 m/s)^2= \frac{50}{23}J$

by the "work-kinetic energy theorem " (work done by the external forces is equals to the net change in the kinetic energy)

work done by the external force is

$W_{external}=KE_{final}-KE_{initial}=\frac{50}{23}J-10J =-\frac{180}{23}J$

Hence the labour or the work was done by the braking force

$\\W_{braking}=\frac{180}{23}J =7.83 J$