Question

a. In $FIGUREEX10.27$, what minimum speed does a $100g$particle need at point A to reach point B?

b. What minimum speed does a $100g$particle need at point B to reach point A?

Short answer

(a) Minimum speed at point A =$7.74m/s.$

(b) Minimum speed at point B =$10m/s$.

Step-by-step solution

Speed : 

It is the change in an object's location with regard to time.

(a) Explanation : 

Particle's mass $m=100gm=0.1kg$,

Formula to determine the kinetic energy of a particle, ,

$KE=\frac{1}{2}m{v}^2$

$KE$= kinetic energy,

$m$= particle's mass,

$v$= particle's velocity.

Consider the graph of potential energy against distance,

The minimal speed necessary by the particle to get from point A to point B is the speed at the top of the graph. The particle's potential energy is $5J$at the top of the graph. Calculate the energy difference between the top of the curve and point A.

$$\begin{gathered} ∆PE=5-2 \\ =3J \end{gathered}$$

The change in potential energy is equal to the change in kinetic energy, according to the law of energy conservation. Hence,

$$\begin{gathered} ∆KE=∆PE \\ =3J \end{gathered}$$

Formula to calculate the minimum speed required at point A,

$∆KE=\frac{1}{2}m{v}_A^2$

Substitute values in equation,

$$\begin{gathered} 3=\frac{1}{2}\times 0.1\times v_A^2 \\ {v}_A=7.74m/s \end{gathered}$$

Hence, the particle's minimum velocity to go from point A to point B is $7.74m/s$.

(b) Explanation : 

The minimal speed necessary by the particle to get from point A to point B is the speed at the top of the graph. The particle's potential energy is $5J$at the top of the graph. Calculate the energy difference between the top of the curve and point B.

$$\begin{gathered} ∆PE=5-0 \\ =5J \end{gathered}$$

The change in potential energy is equal to the change in kinetic energy, according to the law of energy conservation. Hence,

$$\begin{gathered} ∆KE=∆PE \\ =5J \end{gathered}$$

Formula to calculate the minimum speed required at point B,

$∆KE=\frac{1}{2}m{v}_B^2$

Substitute the values,

$$\begin{gathered} 5=\frac{1}{2}\times 0.1\times v_B^2 \\ {v}_B=10m/s \end{gathered}$$

Hence, the particle's minimum velocity to go from point A to point B is$10m/s$.