To solve this problem it is necessary to apply the concepts related to
conservation of energy, for this case manifested through work and kinetic energy.
![W = \Delta KE](https://img.qammunity.org/2020/formulas/physics/college/z4gvf4v42r67zqvc74n6nr6rw586fz5oho.png)
![W = F*d](https://img.qammunity.org/2020/formulas/physics/college/50gbte0p4tsfvnikyw65s19sgeqmmp5xd8.png)
Where,
F= Force (Frictional at this case
)
d= Distance
![\Delta KE = (1)/(2) mv^2](https://img.qammunity.org/2020/formulas/physics/college/y0tnt06h0isfhmrghdgaf01wljsw0jw6sn.png)
Where,
m = mass
v = velocity
Equation both terms,
![F*d = (1)/(2)mv^2](https://img.qammunity.org/2020/formulas/physics/college/f7fwtoep3onwrlaq9w7c80v3thgqpfijyx.png)
![\mu mg *d = (1)/(2)mv^2](https://img.qammunity.org/2020/formulas/physics/college/k6oarclk2a6gw1ov9xvlanoyepyyxqyn02.png)
![\mu g * d = (1)/(2)v^2](https://img.qammunity.org/2020/formulas/physics/college/7l32hi2oxovs0na09a4c69x079v7e1bf2l.png)
![d = (1)/(2) (v^2)/(\mu g)](https://img.qammunity.org/2020/formulas/physics/college/luqedcn4srxnqare2vxo3kiy6umm01yx4j.png)
Replacing with our values we have that
![d = (1)/(2) (25^2)/(0.65*9.8)](https://img.qammunity.org/2020/formulas/physics/college/8hxuks7ry8sw3qffl155s2mse5f9t1tbqj.png)
![d = 49.05m](https://img.qammunity.org/2020/formulas/physics/college/x3kxpm2oersq1umjqpfj58u1xhpwm4xuv9.png)
Therefore the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck is 49.05m