Question

Use work and energy to find an expression for the speed of the block in FIGURE $P10.52$ just before it hits the floor if

(a) the coefficient of kinetic friction for the block on the table is ${\mu }_k$and

(b) the table is friction less

Short answer

(a) The speed of the block just before it hits the floor when the table has coefficient of kinetic friction ${\mu }_k$$\sqrt{\frac{2gh\left(m-{\mu }_{k}M\right)}{m+M}}$

(b) The speed of the block just before it hits the floor when the table is frictionless is$\sqrt{\frac{2mgh}{m+M}}$

Step-by-step solution

Given information (part a)

Given a mass pulley system. The table is either frictionless or has a coefficient of kinetic friction ${\mu }_k$.

Explanation (part a)

From the conservation of energy, the loss of P.E. of the block with mass m is used up in the gain of K.E. of both the blocks and in doing work against friction.

Let m be the mass of the block and v be the speed of the block.

$mgh=\frac{1}{2}(m+M)v^2+{\mu }_{k}Mgh$

$\frac{1}{2}(m+M)v^2=mgh-{\mu }_{k}Mgh$

$(m+M)v^2=2gh\left(m-{\mu }_{k}M\right)$

$v^2=\frac{2gh\left(m-{\mu }_{k}M\right)}{m+M}$

localid="1648546745696" $v=\sqrt{\frac{2gh\left(m-{\mu }_{k}M\right)}{m+M}}$

Given information (part b)

Given a mass pulley system. The table is frictionless.

Explanation (part b)

Speed of the block:

$v=\sqrt{\frac{2gh\left(m-{\mu }_{k}M\right)}{m+M}}$

For a frictionless table, ${\mu }_k=0.$

$\nu =\sqrt{\frac{2mgh}{m+M}}$