Chapter 6, Problem 93P

To determine

The tension on cord using the work –energy theorem, Newton’s law and give your preferred way of solving the system with reason.

Answer to Problem 93P

The tension is $27\text{N}$ and the work-energy theorem is the preferred method because of simple calculations without considering the vector quantities.

Explanation of Solution

Mass of block is $4.0\text{kg}$, length of plane is $8.0\text{m}$, inclination of plane is $15°$, and the distance from the top when cord grasped is $5.0\text{m}$.

Refer the figure 1 given below.

Write the equation for work- energy theorem for the motion of block from top to bottom of the incline.

$W_c+W_{nc}=\Delta K$

Here, the work done by the conservative force is $W_c$, work done by the non-conservative force is $W_{nc}$, and the change in kinetic energy at the top and the bottom of incline.

Rewrite the above equation by considering that the velocity at both top and bottom of incline is zero. This makes the change in kinetic energy also zero.

$W_c+W_{nc}=0$ (I)

Write the relation between $W_c$ and the change in potential energy at the top and bottom of incline.

$W_c=-\Delta U$ (II)

Here, the change in in potential energy at the top and bottom of incline is $\Delta U$.

Write the relation between $W_{nc}$ and tension on cord.

$W_{nc}=-T{d}_2$ (III)

Here, the tension on the cord is $T$, distance from bottom to block when the cord grasped is $d_2$.

Rewrite equation (I) by substituting equation (II) and (III).

$\begin{array}{c}-\Delta U+-T{d}_2=0\\ T=\frac{-\Delta U}{d_2}\end{array}$ (IV)

Write the equation for $\Delta U$.

$\Delta U=mg\left(y_1-y_2\right)$

Here, the mass of block is $m$, gravitational acceleration is $g$, bottom position of incline is $y_1$, and the height of incline is $y_2$.

Write the equation for $y_2$.

$y_2=d\sin \theta$

Here, the length of incline is $d$ and the angle of inclination is $\theta$.

Rewrite the equation for $\Delta U$ by substituting $0\text{m}$ for $y_1$ and the above relation for $y_2$.

$\begin{array}{c}\Delta U=mg\left(0\text{m}-d\sin \theta \right)\\ =-mgd\sin \theta \end{array}$

Rewrite equation (IV) by substituting the above relation for $\Delta U$.

$T=\frac{mgd\sin \theta }{d_2}$ (V)

Refer figure 2 shown below to find $T$ using Newton’s law.

To apply the Newton’s law, motion of block is divided into two parts. First part is the motion of block from the top of incline to until a moment just before the cord is grasped. Note that at block starts from rest at the top of incline.

Second part is the motion is from the point of cord is grasped to till the block comes to rest. Note the first part of motion is a speed-up process and the second part is a speed down process.

Write the equation for net force in x-direction.

$N-mg\cos \theta =0$

Here, the normal reaction force on the block is $N$.

Write the equation for net force in x-direction.

$mg\sin \theta =ma$

Here, the acceleration of block is $a$.

Rewrite the above relation by eliminating $m$.

$g\sin \theta =a$

Write the Newton’s third equation of motion for the block from top just before the cord is grasped.

$v^2-0=2\left(g\sin \theta \right)d_1$ (VI)

Here, the velocity of block just before the cord is grasped is $v$ and the distance from top of incline to the point of block when the cord is grasped is $d_1$.

Write the equation for x-component of force when the cord is grasped.

$-T+mg\sin \theta =m{a}_x$

Here, the x-component of acceleration is $a_x$.

Rewrite the above relation in terms of $a_x$.

$a_x=-\frac{T}{m}+g\sin \theta$ (VII)

Write the Newton’s third equation of motion for the second part of motion. Note that the final velocity of first part of motion is the initial velocity of second part of the motion.

$0-v^2=2a_x{d}_2$

Rewrite the above equation by substituting the equations (VI) and (VII).

$0-2\left(g\sin \theta \right)d_1=2\left(-\frac{T}{m}+g\sin \theta \right)d_2$

Rewrite the above relation in terms of $T$.

$T=mg\sin \theta \left(\frac{d_1+d_2}{d_2}\right)$ (IX)

Write the relation between $d,d_1$ and $d_2$.

$d=d_1+d_2$

Rewrite equation (IX) by substituting the above relation for $d$.

$T=mg\sin \theta \left(\frac{d}{d_2}\right)$

The above equation is same as that of equation (V).Thus; both methods will give the same relation for $T$.

Conclusion:

Substitute $4.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $15°$ for $\theta$, $8.0\text{m}$ for $d$, and $3.0\text{m}$ in the above equation to find $T$.

$\begin{array}{c}T=\left(4.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(\sin 15°\right)\left(\frac{8.0\text{m}}{3.0\text{m}}\right)\\ =27\text{N}\end{array}$

The method using work –energy theorem is preferred. This is because of the reasons like less calculations, no need of fee body diagram. No need of concern about direction of vector quantities, and need to split the motion into two parts.

Therefore, the tension is $27\text{N}$ and the work-energy theorem is the preferred method because of simple calculations without considering the vector quantities.