Question

the displacement of the package relative to the belt as the belt comes to stop. (g= $9.8m/s^2$).

Answer 1

Initial velocity of belt $u=0m/s$

Velocity of belt after 1.3s $v=u+at=0+2\times 1.3=2.6m/s$

Distance traveled in 1.3 sec is $s=\frac{a{t}^2}{2}=\frac{2\times {1.3}^2}{2}=1.69m$

Now the total displacement after it stops is $d=2.2m$

So distance traveled during deceleration is $s_2=d-s=2.2-1.69=0.51m$

So let the deceleration be $a_2$.

Using $(v^{\prime }{)}^2-v^2=2(-a_2)s_2$ where final velocity of belt after deceleration $v^{\prime }=0m/s$

Or $0-{2.6}^2=-2\times 0.51\times a_2$

$⟹a_2=6.63m/s^2$

Time taken to stop

$t_2=\frac{v}{a_2}=0.4s$

Since the maximum kinetic friction is $0.25$, so the maximum acceleration of package is ${\mu }_{k}g=0.25g=2.5m/s^2$, which is greater than the acceleration of belt ($2m/s^2$). So it will accelerate at $a_p=2m/s^2$

Since the coefficient of kinetic friction is $0.25$, so the maximum

deceleration of package is ${\mu }_{k}g=0.25g=2.5m/s^2$, which is greater than the deceleration of belt ($6.63m/s^2$). So it will decelerate at $a_r=2.5m/s^2$

So velocity of package after 1.3 sec is $v_p=2\times 1.3=2.6m/s$

Distance traveled in 1.3 sec is $s_p=\frac{a{t}^2}{2}=\frac{2\times {1.3}^2}{2}=1.69m$

When the system starts decelerating, distance traveled before belt stops

$s_r=v_p{t}_2-\frac{a_r{t}_2^2}{2}=2.6\times .4-\frac{{\mathrm{2.50.4}}^2}{2}$

$s_r=0.84m$

So relative to belt the displacement of package is $s_r-s_2=0.84-0.51=0.33m$