On the edge of a cart of mass $M$ and length $l$ standing on a smooth table there is a small cube of mass $m$. The cube is pushed with a constant force $F$ directed horizontally to the opposite edge of the cart. In what time will the cube reach the opposite edge of the cart? The coefficient of friction between the cube and the cart is $\mu$. The cube does not roll over.
Decision:
Considering the forces acting on the cube and the cart and examining the equations of motion of the cube and cart, we arrive at the following result.
If $F \leq \mu mg \frac{m+M}{M}$, then the load will not move at all relative to the cart.
If $F> \mu mg \frac{m+M}{M}{M}$, the load slips, and its acceleration relative to the table
$a=\frac{F}{m}-\mu g$.
Only the force of friction acts on the cart in the horizontal direction, so its acceleration relative to the table is
$a_{a}=\mu \frac{m}{M}{M} g$.
The load moves relative to the cart with acceleration
$a_{0}=a-a_{1}=\frac{F}{m}-\mu g \left ( 1+\frac{m}{M} \right)$
and will reach the edge of the cart in the time
$t=\sqrt{\frac{2l}{a_{0}}}=\sqrt{\frac{2l}{ \frac{F}{m}-\mu g \left ( 1+\frac{m}{M} \right )}}$
Decision:
Considering the forces acting on the cube and the cart and examining the equations of motion of the cube and cart, we arrive at the following result.
If $F \leq \mu mg \frac{m+M}{M}$, then the load will not move at all relative to the cart.
If $F> \mu mg \frac{m+M}{M}{M}$, the load slips, and its acceleration relative to the table
$a=\frac{F}{m}-\mu g$.
Only the force of friction acts on the cart in the horizontal direction, so its acceleration relative to the table is
$a_{a}=\mu \frac{m}{M}{M} g$.
The load moves relative to the cart with acceleration
$a_{0}=a-a_{1}=\frac{F}{m}-\mu g \left ( 1+\frac{m}{M} \right)$
and will reach the edge of the cart in the time
$t=\sqrt{\frac{2l}{a_{0}}}=\sqrt{\frac{2l}{ \frac{F}{m}-\mu g \left ( 1+\frac{m}{M} \right )}}$