Question

Part AA 2.75 kg bale of hay with an initial speed of 1.75 m/s slides along the ground, moving 40.0 cm before coming to rest. Find the coefficient of kinetic friction between the ground and the bale of hay.Part BIf the bale of hay had been traveling twice as fast, what distance would it have skidded, under the same conditions?It would have skidded 4 times farther.It would have skidded one half as far.It would have skidded one-fourth as far.It would have skidded twice as far.It is impossible to tell from the information given.

Answer 1

$A)\mu=0.363$

B)It would have skidded 4 times farther.

Explanation

the work is the change in energy that results from appliyng a force , in this case, the force is the force of friction

$\text{work}=\text{change in energy= Final energy-initial energy}$

Step 1

find the work done:

Let

$\begin{gathered} \text{mass}=2.75\text{ kg} \\ v_1=1.75\text{ }\frac{m}{s} \\ \text{distance}=\text{ 40 cm =0.4 m} \\ v_f=0\text{ (rest) } \end{gathered}$

so

$\begin{gathered} \Delta energy \\ E_f-E_i=\frac{1}{2}mv^2-\frac{1}{2}mv^2_2 \\ E_f-E_i=\frac{1}{2}mv^2-0 \\ E_f-E_i=\frac{1}{2}(2.75)(1.75)^2 \\ E_f-E_i=4.21\text{ Joules} \\ E_f-E_i=4.2\text{ Jouls} \end{gathered}$

hence, the work and applied force are:

$\begin{gathered} W=\text{ force}\cdot dis\tan ace \\ W=4.2\text{ joules, } \\ \text{hence} \\ 4.2=F\cdot d \\ \text{led d= 0.4 and replace} \\ \frac{4.2}{d}=F \\ F=10.52\text{ Newtons} \end{gathered}$

Step 2

$\begin{gathered} \text{The force of friction is given by:} \\ Ff=\mu\cdot N \\ \text{wherre }\mu\text{ is the coeffictient of frictin ans N is the normal, so} \\ \mu=\frac{Ff}{N} \\ \text{and remember that} \\ N=mg \\ so \\ \mu=\frac{Ff}{mg} \end{gathered}$

now, replace

$\begin{gathered} \mu=\frac{Ff}{mg} \\ \mu=\frac{10.52}{(2.75\operatorname{kg})(1.75\frac{m}{s})} \\ \mu=0.363 \end{gathered}$

therefotre, the cofficient of kinetick friction is 0.363

Step 3

Part B

if the bale of hay had been traveling twice as fast

then let

$v_a=2v$

so

a)

$\begin{gathered} \Delta energy \\ E_f-E_i=\frac{1}{2}m(2v)^2-\frac{1}{2}mv^2_2 \\ E_f-E_i=\frac{4}{2}mv^2-0 \\ E_f-E_i=2(2.75)(1.75)^2 \\ E_f-E_i=16.84\text{ Joules} \\ E_f-E_i=16.8\text{ Jouls} \end{gathered}$

now, replace in the formula

as the force is the same, replace

$\begin{gathered} W=\text{ force}\cdot dis\tan ace \\ W=16.8 \\ \text{hence} \\ 4.2\cdot4=F\cdot d \\ d=4(\frac{4.2}{F}) \\ \end{gathered}$

so, we can conclude that

It would have skidded 4 times farther.

I hope this helps you