Answer:
The answer is "[tex]2.41 \times 10^3[/tex]"
Explanation:
Given:
[tex]m_i = 2000 \ kg \\\\v_i= 4.1 \ \frac{m}{s} \\\\v_f = 3.4 \ \frac{m}{s} \\[/tex]
Using formula:
[tex]\to m_iv_i = m_fv_f \\\\\to m_f= \frac{m_iv_i}{v_f}[/tex]
[tex]p_i, p_f[/tex] = system initial and final linear momentum.
[tex]V_i, v_f[/tex] = system original and final linear pace.
[tex]m_i[/tex] = original weight of the car freight.
[tex]m_f[/tex]= car's maximum weight
[tex]= \frac{ 2000 \times 4.1}{3.4}\\\\= \frac{ 8.2\times 10^3}{3.4}\\\\= 2.41 \times 10^3[/tex]
[tex]\boxed{m_f = 2.41 \times 10^3}[/tex]
