Question

Review: As shown in Figure P8.46, a light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects $m_1$, $3.50-kg$a block originally at rest on the horizontal table at a height $h=1.20m$above the floor, to $m_2$, a hanging $1.90-kg$ block originally a distance $d=0.900m$above the floor. Neither the surface of the table nor its edge exerts a force of kinetic friction. The blocks start to move from rest. The sliding block $m_1$ is projected horizontally after reaching the edge of the table. The hanging block $m_2$ stops without bouncing when it strikes the floor. Consider the two blocks plus the Earth as the system. (a) Find the speed at which leaves the edge of the table. (b) Find the impact speed of $m_1$on the floor. (c) What is the shortest length of the string so that it does not go taut while $m_1$is in flight? (d) Is the energy of the system when it is released from rest equal to the energy of the system just before $m_1$strikes the ground? (e) Why or why not?

Short answer

(a) the speed at is $v=2.49m/s$.

(b) the impact speed is $v_d=5.45m/s$.

(c) the shortest length of the string is $1.23m$.

(d) No, the energy will not be equal.

(e) Some of the kinetic energy of $m_2$ is transformedto sound and some is transformed to internal energy and stored in $m_1$.

Step-by-step solution

Step 1:

The law of conservation of energy states that the total energy of an isolated system is conserved, which can be written as

${\mathbf{\Delta E}}_{\mathrm{system}}\mathbf{=}\mathbf{0}$

$\mathbf{\Delta K}\mathbf{+}\mathbf{\Delta U}\mathbf{=}\mathbf{0}$

The potential energy of the system is

$\mathbf{U}\mathbf{=}\mathbf{mgR}$

The kinetic energy of the system is

$\mathbf{K}\mathbf{=}\frac{1}{2}{\mathbf{mv}}^2$

where

$U=$Potential energy

$K=$Kinetic energy

$v=$Speed of body

$g=$Gravitational acceleration

$R=$Displacement in meter

$m=$ Mass of body.

Step 2:Stating given data

Mass of the body placed on the table$m_1=3.5\text{\hspace{0.17em}kg}$

Mass of the body hanging from the table$m_2=1.9\text{\hspace{0.17em}kg}$

The height at which body is hanging $d=0.90\text{\hspace{0.17em}m}$

The height of the table $h=1.2\text{\hspace{0.17em}m}$.

Calculation

Part (a):

Using law of conservation of energy-

$\begin{array}{l}{m}_{2}gy=\frac{1}{2}(m_1+m_2)v^2\\ v=\sqrt{\frac{2m_{2}gy}{m_1+m_2}}\\ v=\sqrt{\frac{2(1.90kg)(9.80m/s^2)(0.900m)}{5.40kg}}\\ v=2.49m/s\end{array}$

Part (b):

For the$3.50kg$ block from when the string goes slack until just before the block hits the floor, conservation of energy gives

$\begin{array}{l}\frac{1}{2}\left(m_2\right)v^2+m_{2}gy=\frac{1}{2}\left(m_2\right)v_d^2\\ v_d=\sqrt{2gy+v^2}\\ v_d=\sqrt{2(9.8m/s)(1.20m)+(2.49m/s^2)}\\ v_d=5.45m/s\end{array}$

Part (c):

The time taken by $3.50kg$ block to reach the floor.

From second equation of motion, if initial velocity is zero, we have-

$\begin{array}{l}y=\frac{1}{2}g{t}^2\\ t=\sqrt{\frac{2y}{g}}\end{array}$

$\begin{array}{c}t=\sqrt{\frac{2\left(1.2\text{\hspace{0.17em}m}\right)}{(9.8m/s^2)}}\\ =0.495s\end{array}$

At the time of impact, the horizontal component of displacement, will be-

$\begin{array}{c}x=v_{d}t\\ =(2.49m/s)(0.495s)\\ =1.23m\end{array}$

So, the shortest length of the string is $1.23m$

Part (d):

No, the energy of the system, when it is released from rest is not equal to the energy of the system just before $m_1$ strikes the ground.

Part (e):

Some of the kinetic energy of $m_2$ is wasted as sound and some is transformed to internal energy of $m_1$and the floor.