Question

A dinner plate on a tablecloth,with its center $0.3$ m from the edge of the table.The tablecloth is suddenly yanked with a constant acceleration of $9.2m/s^{\wedge }2$The coefficient of friction $\left(u\right)=0.75$.Find $\left(1\right)$ the acceleration $\left(2\right)$ the velocity and $\left(3\right)$ the distance of the plate from the edge of the table.When the edge of the tablecloth passes under the center of the plate $\left[7.35m/s^{\wedge }2,4.26m/s,1.54m\right]$

Answer 1

$\mu =0.75$

$\left(1\right)$ maximum value of friction force that can act on plate :

$f_l=\mu mg$

$ma=\mu mg$

$a=\mu g$

$a=0.75\times 9.8$

$a=7.35m{s}^{-2}$

$\left(2\right)t$ be the time when tablecloth pass under plate

$0.3=\frac{(9.2-7.35)t^2}{2}$

$t=\sqrt{\frac{0.6}{1.85}}$

$v=at$

$v=7.35\times \sqrt{\frac{0.6}{1.85}}$

$v=4.18m/s$

$\left(3\right)$ $s+0.3$

$s=\frac{a{t}^2}{2}$

$s=\frac{7.35}{2}\times \frac{0.6}{1.85}$

$s=1.19$

$s+0.3=1.49m$