Question

A 1.50 kg book is sliding along a rough horizontal surface. At point A it is moving at 3.21 m/s , and at point B it has slowed to 1.25 m/s .

a.How much work was done on the book between A and B?

b.If -0.750 of work is done on the book from B to C, how fast is it moving at point C? in m/s

c. How fast would it be moving at C if 0.750 J of work were done on it from B to C? in m/s

Answer 1

Concepts and reason

The concepts used to solve this problem is conservation of the energy principle and work.

Initially, the conservation of the energy principle the work done by the system is calculated. From the expression of the work done the change in kinetic energy is equal to the work done by the object.

Fundamentals

The expression of the change in kinetic energy is equal to,

$\Delta KE = K{E_f} - K{E_i}$

Here,$\Delta KE$is the change in kinetic energy,$K{E_f}$is the final kinetic energy and $K{E_i}$is the initial kinetic energy.

From the definition of the work the expression of the work done is equal to,

$W = - \Delta KE$

Here,$W$is the work done.

(a)

To determine the value of work done between A and B, the expression of the work done is equal to,

${W_t} = {K_B} - {K_A}$

Substitute $\frac{1}{2}m{V_B}^2$for${K_B}$and$\frac{1}{2}m{V_A}^2$for ${K_A}$

$\begin{array}{c}\\{W_t} = \frac{1}{2}m{V_B}^2 - \frac{1}{2}m{V_A}^2\\\\ = \frac{1}{2}m\left( {{V_B}^2 - {V_A}^2} \right)\\\end{array}$

Substitute$1.50{\rm{ kg}}$for$m$,$3.21{\rm{ m/s}}$for${V_A}$and $1.25{\rm{ m/s}}$fro${V_B}$ in the above expression of the work done,

$\begin{array}{c}\\{W_t} = \frac{1}{2}\left( {1.50{\rm{ kg}}} \right)\left( {{{\left( {1.25{\rm{ m/s}}} \right)}^2} - {{\left( {3.21{\rm{ m/s}}} \right)}^2}} \right)\\\\ = - 6.56{\rm{ J}}\\\end{array}$

(b)

To determine the out how fast the book is moving at point C if the work was done is -0.750 J, the expression of the work in equal to,

${W_t} = \frac{1}{2}m\left( {{V_C}^2 - {V_B}^2} \right)$

Substitute $- 0.75{\rm{ J}}$for${W_t}$,$1.25{\rm{ m/s}}$for${V_B}$and $1.50{\rm{ kg }}$for$m$in the above expression of the work done.

$\begin{array}{c}\\ - 0.75{\rm{ J}} = \frac{1}{2}\left( {1.50{\rm{ kg }}} \right)\left( {{V_C}^2 - {{\left( {1.25{\rm{ m/s}}} \right)}^2}} \right)\\\\\frac{{ - 1.50{\rm{ J}}}}{{\left( {1.50{\rm{ kg }}} \right)}} = \left( {{V_C}^2 - {{\left( {1.25{\rm{ m/s}}} \right)}^2}} \right)\\\end{array}$

Rearrange the above expression in terms of the velocity${V_c}$.

$\begin{array}{c}\\{V_C} = \sqrt {{{\left( {1.25{\rm{ m/s}}} \right)}^2} - \frac{{1.50{\rm{ J}}}}{{\left( {1.50{\rm{ kg }}} \right)}}} \\\\ = 0.75{\rm{ m/s}}\\\end{array}$

The expression of the work done in the book which is moving at point C if the work done is -0.750 J is given by,

${W_t} = \frac{1}{2}m\left( {{V_C}^2 - {V_B}^2} \right)$

Substitute $1.50{\rm{ kg}}$for$m$,$1.25{\rm{ m/s}}$for${V_B}$and $0.750{\rm{ J}}$for${W_t}$in the above expression of the work.

${W_t} = \frac{1}{2}m\left( {{V_C}^2 - {V_B}^2} \right)$

Find out how fast the book is moving at point C if the work done is -0.750 J is given by

$\begin{array}{l}\\0.750{\rm{ J}} = \left( {\frac{1}{2}} \right)\left( {1.50{\rm{ kg}}} \right)\left( {{V_C}^2 - {{\left( {1.25{\rm{ m/s}}} \right)}^2}} \right)\\\\\frac{{\left( {0.750{\rm{ J}}} \right)\left( 2 \right)}}{{\left( {1.50{\rm{ kg}}} \right)}} = {V_C}^2 - {\left( {1.25{\rm{ m/s}}} \right)^2}\\\end{array}$

Rearrange the above expression in terms of the velocity${V_C}$.

$\begin{array}{c}\\{V_C} = \sqrt {{{\left( {1.25{\rm{ m/s}}} \right)}^2} + \frac{{\left( {0.750{\rm{ J}}} \right)\left( 2 \right)}}{{\left( {1.50{\rm{ kg}}} \right)}}} \\\\ = 1.60{\rm{ m/s}}\\\end{array}$

Ans: Part a

The work done on the book between ${\rm{A and B}}$is equal to$- 6.56{\rm{ J}}$.

Part b

The book is moving at a speed of ${\rm{0}}{\rm{.75 m/s}}$ at point C.

Part c

The book is moving at a speed of $1.60{\rm{ m/s}}$ at point C.