Question

A 1.50 kg book is sliding along a rough horizontal surface. At point A it is...

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
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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,

ΔKE=KEfKEi\Delta KE = K{E_f} - K{E_i}

Here,ΔKE\Delta KEis the change in kinetic energy,KEfK{E_f}is the final kinetic energy and KEiK{E_i}is the initial kinetic energy.

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

W=ΔKEW = - \Delta KE

Here,WWis the work done.

(a)

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

Wt=KBKA{W_t} = {K_B} - {K_A}

Substitute 12mVB2\frac{1}{2}m{V_B}^2forKB{K_B}and12mVA2\frac{1}{2}m{V_A}^2for KA{K_A}

Wt=12mVB212mVA2=12m(VB2VA2)\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}

Substitute1.50kg1.50{\rm{ kg}}formm,3.21m/s3.21{\rm{ m/s}}forVA{V_A}and 1.25m/s1.25{\rm{ m/s}}froVB{V_B} in the above expression of the work done,

Wt=12(1.50kg)((1.25m/s)2(3.21m/s)2)=6.56J\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,

Wt=12m(VC2VB2){W_t} = \frac{1}{2}m\left( {{V_C}^2 - {V_B}^2} \right)

Substitute 0.75J - 0.75{\rm{ J}}forWt{W_t},1.25m/s1.25{\rm{ m/s}}forVB{V_B}and 1.50kg1.50{\rm{ kg }}formmin the above expression of the work done.

0.75J=12(1.50kg)(VC2(1.25m/s)2)1.50J(1.50kg)=(VC2(1.25m/s)2)\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 velocityVc{V_c}.

VC=(1.25m/s)21.50J(1.50kg)=0.75m/s\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,

Wt=12m(VC2VB2){W_t} = \frac{1}{2}m\left( {{V_C}^2 - {V_B}^2} \right)

Substitute 1.50kg1.50{\rm{ kg}}formm,1.25m/s1.25{\rm{ m/s}}forVB{V_B}and 0.750J0.750{\rm{ J}}forWt{W_t}in the above expression of the work.

Wt=12m(VC2VB2){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

0.750J=(12)(1.50kg)(VC2(1.25m/s)2)(0.750J)(2)(1.50kg)=VC2(1.25m/s)2\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 velocityVC{V_C}.

VC=(1.25m/s)2+(0.750J)(2)(1.50kg)=1.60m/s\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 AandB{\rm{A and B}}is equal to6.56J - 6.56{\rm{ J}}.

Part b

The book is moving at a speed of 0.75m/s{\rm{0}}{\rm{.75 m/s}} at point C.

Part c

The book is moving at a speed of 1.60m/s1.60{\rm{ m/s}} at point C.

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