Given information:
[tex]F=2.5*10^3 N=2500 N\\v_0=0\\v_f=48 \frac{km}{h}[/tex]
Convert the units of velocity into something more useful for this problem:
[tex](\frac{48 km}{1 h}) (\frac{1 h}{3600 s} )(\frac{1000 m}{1 km} )=\frac{48000 m}{3600 s} =\frac{40}{3} \frac{m}{s}[/tex]
We know [tex]F=ma[/tex]. Solving for mass gives [tex]m=\frac{F}{a}[/tex]
Now use a kinematic equation to find acceleration in terms of known information.
[tex]v=v_0+at\\a=\frac{v-v_0}{t} [/tex]
Plug in and solve.
[tex]m=\frac{F}{a} =\frac{F}{\frac{v-v_0}{t}}\\m=\frac{2500 N}{\frac{\frac{40}{3}\frac{m}{s} -0\frac{m}{s} }{5.0 s}}=\frac{1875}{2} kg=937.5 kg[/tex]
This gives us a mass of 937.5 kg.
However, the least amount of significant figures we had was 2, so with correct significant figures, this would be 940 kg.
