All Solutions
Page 278: Assessment
On that object acts gravitational force downwards and there is reaction from surface to gravitational force with same magnitude but in opposite direction.
Both forces are perpendicular to the motion and they do not do any work.
$$
W=Fscosphi
$$
where $phi$ is angle between direction of motion and force.
$$
P=frac{W}{t}
$$
W=Fscosphi
$$
$$
P=frac{W}{t}
$$
$$P=frac{W}{t}$$
and that work is equal to:
$$W=Fs$$
unit for force is N and it is equal to $rm kgm/s^2$, that means that unit for work is equal to $rm kgm^2/s^2$, now unit for power is equal to:
$$rm P= kgm^2/s^3$$
This means that **it is not possible** because energy can not come from nothing and we cannot get more energy than was input into the machine.
$$
W=Fs
$$
in this case force is weight and distance is height.
$$
W_1=420cdot200=84000,,rm J
$$
$$
W_2=210cdot400=84000,,rm J
$$
WE got that it requires equal work.
upstairs needs positive work; carrying
it back down is negative work. The work
done in both cases is equal and opposite
because the distances are equal and
opposite. The student might arrange the
payments on the basis of the time it
takes to carry paper, whether up or
down, not on the basis of work done.
$$
W=Fdcostheta
$$
We can use equation for work to answer this question:
$$
W=Fd
$$
in this case force is equal to weight and distance is equal to height.
$$
W=mgh
$$
We know that they have the same mass and that they climb the same stairs that means that work is $textbf{the same}$ for both of them.
We can use equation for power to answer this question:
$$
P=frac{W}{t}
$$
We know from part $textbf{a}$ that work is equal but in this case time is different and it is the $textbf{first person}$ that has generated more power because it needed less time to climb.
b) first person
$$
P=Fvcostheta
$$
$$
P=frac{W}{t}
$$
now we can write work in terms of force and distance:
$$
W=Fdcostheta
$$
when we connect those equations we get:
$$
P=frac{Fdcostheta}{t}
$$
now we know that velocity is equal to distance per time:
$$
v=frac{d}{t}
$$
When we use that in previous equation we get:
$$
P=Fvcostheta
$$
$$
IMA=frac{d_e}{d_r}
$$
From this equation we can see that we have to increase ratio between input displacement and output displacement.
$$
W=mgh
$$
$$
W=150cdot9.81cdot8
$$
$$
boxed{W=11772,,rm J}
$$
W=11772,,rm J
$$
$$
W=mghrightarrow m=frac{W}{gh}
$$
$$
m=frac{176}{9.81cdot0.3}
$$
$$
boxed{m=59.8,,rm kg}
$$
m=59.8,,rm kg
$$
$$
W=mgh
$$
$$
W=84cdot9.81cdot1.2
$$
$$
boxed{W=988.8,,rm J}
$$
W=988.8,,rm J
$$
$$
W=Fdrightarrow F=frac{W}{d}
$$
$$
F=frac{2.2 cdot10^{5}}{8}
$$
$$
boxed{F=27500,,rm N}
$$
F=27500,,rm N
$$
much work is done against friction by the car. It is given that car needs a 551 N of force to balance frictional force, and distance between Columbus and Cincinnati is 161 km.
$$
W=Fd
$$
$$
W=551cdot161cdot10^{3}
$$
$$
boxed{W=887cdot10^{5},,rm J}
$$
W=887cdot10^{5},,rm J
$$
$$
P=frac{W}{t}
$$
To get work we can use equation:
$$
W=Fd
$$
when we put those equations together we get:
$$
P=frac{Fd}{t}
$$
$$
P=frac{15cdot251}{30}
$$
$$
boxed{P=125.5,,rm W}
$$
P=125.5,,rm W
$$
$$
W=mgh
$$
$$
W=2.2cdot9.81cdot0.35
$$
$$
boxed{W=7.55,,rm J}
$$
W=7.55,,rm J
$$
$$
W=Fd
$$
and we can calculate power with equation:
$$
P=frac{W}{t}
$$
When we put numbers we get:
$$
W=300cdot30
$$
$$
boxed{W=9000,,rm J}
$$
When we put numbers in we get:
$$
P=frac{9000}{3}
$$
$$
boxed{P=3000,,rm W}
$$
b) $P=3000,,rm W$
$$
F_h=Fcosphi
$$
In this case total work is equal to kinetic energy, to get kinetic energy we can use equation for centripetal force:
$$
F_{cp}=mfrac{v^2}{r}rightarrow v=sqrt{frac{F_{cp}r}{m}}
$$
in this case centripetal force is equal to horizontal force.
Now we can get kinetic energy:
$$
E_k=frac{mv^2}{2}
$$
And kinetic energy is equal to work.
When we put those equations together we get:
$$
W=frac{Fcosphicdot r}{2}
$$
$$
W=frac{38cos42^ocdot25}{2}
$$
$$
boxed{W=353,,rm J}
$$
W=353,,rm J
$$
$$
W=Fdcosphi
$$
$$
W=88cdot1.2cdot10^{3}cos41^o
$$
$$
boxed{W=79697,,rm J}
$$
W=79697,,rm J
$$
textbf{Givens: }
$$
$$
begin{align*}
&m=17text{ kg}\\
&x=20text{ m}\\
&W=1210text{ J}\\
&F=75text{ N}
end{align*}
$$
textbf{Analysis: }
$$
* The work done by the rope is equivalent to the pulling force multiplied by the distance multiplied by the cosine angel of the rope:
$$
W=Fxcos(theta)
$$
* So, to determine $theta$, we should divide the work by the force multiplied by distance and get the inverse of the $cos$ angle for the result:
$$
theta=cos^{-1}(dfrac{W}{Fx})
$$
text{color{default}
item Determine the angle of the rope:
begin{align*}
theta&=cos^{-1}(dfrac{W}{Fx})\\
&=cos^{-1}(dfrac{1210}{75 times 20})\\
&=cos^{-1}(0.14)\\
&=36.23^{circ}
end{align*}
item So, the angle of the rope is:
$$boxed{theta=36.23^{circ}}$$
}
$$
theta=36.23^{circ}
$$
$$
P=frac{W}{t}
$$
Work is equal to change in potential energy:
$$
W=mgh=mgdsinphi
$$
Now we have:
$$
P=frac{mgdsinphi}{t}
$$
$$
P=frac{120cdot9.81cdot12sin21^o}{2.5}
$$
$$
boxed{P=2025,,rm W}
$$
P=2025,,rm W
$$
$$
W=Fd=Ffrac{h}{sin30^o}
$$
$$
W=225cdotfrac{1.15}{sin30^o}
$$
$$
boxed{W=517.5,,rm J}
$$
W=517.5,,rm J
$$
$$
W=F_gh=F_gdsinphi
$$
$$
W=4.2cdot10^{3}cdot3.5sin30^o
$$
$$
boxed{W=7350,,rm N}
$$
W=7350,,rm N
$$
$$
W=Fdcosphi
$$
$$
W=225cdot65.3cos35^o
$$
$$
boxed{W=12035,,rm J}
$$
W=12035,,rm J
$$
$$
W=mgh=mgdsinphi
$$
$$
W=52cdot9.81cdot227sin31^o
$$
$$
boxed{W=59640,,rm J}
$$
W=59640,,rm J
$$
$$
W=Fdcosphirightarrow d=frac{W}{Fcosphi}
$$
To get work we can use equation for power:
$$
P=frac{W}{t}rightarrow W=Pt
$$
when we put those equations together we get:
$$
d=frac{Pt}{Fcosphi}
$$
$$
d=frac{64.6cdot90}{115cdotcos22.5^o}
$$
$$
boxed{d=54.7,,rm m}
$$
d=54.7,,rm m
$$
$$
W=W_1+W_2+W_3=F_1d_1+F_2d_2+F_3d_3
$$
When we put numbers in we get:
$$
W=20cdot5+35cdot12+10cdot8
$$
$$
boxed{W=600,,rm J}
$$
W=600,,rm J
$$
To calculate work we can use equation:
$$
W=Fd
$$
$$
W=400cdot2
$$
$$
boxed{W=800,,rm J}
$$
In this case work is equal to change in potential energy:
$$
W=mgh
$$
$$
W=60cdot9.81cdot1
$$
$$
boxed{W=588.6,,rm J}
$$
b) $W=588.6,,rm J$
$$
P=Fv
$$
$$
P=6cdot10^{3}cdot15
$$
$$
boxed{P=9cdot10^{4},,rm W}
$$
P=9cdot10^{4},,rm W
$$
To calculate the slope we have to pick two point on the graph and put them in equation:
$$
k=frac{y_2-y_1}{x_2-x_1}
$$
and to show that $F=kd$ we have to put any value of $d$ and if we get the same result as on the graph it is shown.
$$
k=frac{5-0}{0.2-0}
$$
$$
boxed{k=25,,rm N/m}
$$
If we put $d=0.08,,rm m$ we get:
$$
F=25cdot0.08=2,,rm N
$$
We can se that it is the same as in graph.
To find amount of work we have to calculate area under the graph:
$$
W=A
$$
$$
W=frac{1}{2}cdot0.2cdot5
$$
$$
boxed{W=0.5,,rm J}
$$
We can see that formula for area under the graph is:
$$
A=frac{1}{2}Fd
$$
And we know that magnitude of force can be calculated as:
$$
F=kd
$$
when we put that in first equation we get:
$$
A=frac{1}{2}kd^2
$$
And because area under the graph is work:
$$
W=frac{1}{2}kd^2
$$
b) $W=0.5,,rm J$
$$
W=A_s+A_t
$$
$$
W=0.16cdot3+frac{1}{2}cdot0.16cdot4
$$
$$
boxed{W=0.8,,rm J}
$$
W=0.8,,rm J
$$
We can calculate work with equation:
$$
W=Fdcosphi=Fdfrac{l}{d}=Fl
$$
$$
W=85cdot4
$$
$$
boxed{W=340,,rm J}
$$
Work done by gravity is equal to change in potential energy:
$$
W=F_gh
$$
$$
W_g=-93cdot3
$$
ž
$$
boxed{W_g=-279,,rm J}
$$
Minus sign is because gravitational force is in opposite direction.
We can calculate total work done by friction with equation:
$$
W_f=F_fd=F_Nmu d
$$
In this case normal force is equal to:
$$
F_N=F_gcosphi+Fsinphi=F_gfrac{l}{d}+Ffrac{h}{d}
$$
Now we get:
$$
W_f=left(F_gfrac{l}{d}+Ffrac{h}{d}right)mu d=left(F_gl+Fhright)mu
$$
$$
W_f=-(93cdot4+85cdot3)cdot0.2
$$
$$
boxed{W_f=-125.4,,rm J}
$$
b) $W_g=-279,,rm J$
c) $W_f=-125.4,,rm J$
Work is equal to difference in potential energy:
$$
W=mgh=rho Vgh
$$
$$
W=820cdot0.55cdot9.81cdot25
$$
$$
boxed{W=110608,,rm J}
$$
Power is equal to work per time:
$$
P=frac{W}{t}
$$
$$
P=frac{110608}{35}
$$
$$
boxed{P=3160,,rm W}
$$
b) $P=3160,,rm W$
$$
P=frac{W}{t}
$$
We can find work as difference in potential energy:
$$
W=mgh=mglsinphi
$$
now we have:
$$
P=frac{mglsinphi}{t}
$$
$$
P=frac{15cdot25cdot9.81cdot12cdotsin30^o}{60}
$$
$$
boxed{P=368,,rm W}
$$
P=368,,rm W
$$
$$
P=Fvrightarrow F=frac{P}{v}
$$
$$
F=frac{48cdot10^{3}}{frac{76}{3.6}}
$$
$$
boxed{F=2274,,rm N}
$$
F=2274,,rm N
$$
Work is equal to area under the graph:
$$W=A$$
$$W=frac{1}{2}cdot20cdot2+frac{1}{2}cdot30cdot1+20cdot5+30cdot4$$
$$boxed{W=255,,rm J}$$
Power is defined as work per time:
$$P=frac{W}{t}$$
$$P=frac{255}{2}$$
$$boxed{P=127.5,,rm W}$$
b) $P=127.5,,rm W$
If this is ideal machine then work don by man on machine is equal to work done by machine on piano:
$$F_gd=Flrightarrow F=frac{F_gd}{l}$$
$$F=frac{1200cdot5}{20}$$
$$boxed{F=300,,rm N}$$
We calculated in part **a** that for ideal machine effort force is $300 ,,rm N$, difference in those forces is equal to force needed to balance the friction force.
$$F_f=F_e-F$$
$$F_f=340-300$$
$$boxed{F_f=40,,rm N}$$
We can calculate output work using force on piano and distance we moved piano:
$$W_{out}=F_g d$$
$$W_{out}=1200cdot5$$
$$boxed{W_{out}=6000,,rm J}$$
We can calculate input work using force pulley and length of pulled rope:
$$W_{in}=F_el$$
$$W_{in}=340cdot20$$
$$boxed{W_{in}=6800,,rm J}$$
Mechanical advantage is defined as:
$$MA=frac{F_r}{F_e}$$
$$MA=frac{1200}{340}$$
$$boxed{MA=3.53}$$
b) $F_f=40,,rm N$
c) $W_{out}=6000,,rm J$
d) $W_{in}=6800,,rm J$
e) $MA=3.53$
$$e=frac{W_o}{W_i}cdot100$$
We can get output work with equation:
$$W_o=mgd$$
Together we get:
$$e=frac{mgd}{W_i}cdot100rightarrow W_i=frac{mgd}{e}cdot100$$
$$W_i=frac{18cdot9.81cdot0.5}{90}cdot100$$
$$boxed{W_i=98.1,,rm J}$$
Ideal mechanical advantage is defined as the ratio of the distances moved:
$$IMA=frac{d_e}{d_r}$$
$$IMA=frac{3.9}{0.975}$$
$$boxed{IMA=4}$$
The mechanical advantage is the ratio of resistance force to effort force:
$$MA=frac{F_r}{F_e}$$
$$MA=frac{1345}{375}$$
$$boxed{MA=3.59}$$
Efficiency is defined as mechanical advantage per ideal mechanical advantage:
$$e=frac{MA}{IMA}cdot100%$$
$$e=frac{3.59}{4}cdot100%$$
$$boxed{e=89.7%}$$
b) $MA=3.59$
c) $e=89.7%$
Ideal mechanical advantage is defined as the ratio of the distances moved:
$$IMA=frac{d_e}{d_r}$$
$$IMA=frac{40}{10}$$
$$boxed{IMA=4}$$
The mechanical advantage is the ratio of resistance force to effort force:
$$MA=frac{F_r}{F_e}$$
Here resistance force is equal to weight:
$$F_r=mg$$
that gives us:
$$MA=frac{mg}{F_e}$$
$$MA=frac{0.5cdot9.81}{1.4}$$
$$boxed{MA=3.5}$$
Efficiency is defined as mechanical advantage per ideal mechanical advantage:
$$e=frac{MA}{IMA}cdot100%$$
$$e=frac{3.5}{4}cdot100%$$
$$boxed{e=87.6%}$$
b) $MA=3.5$
c) $e=87.6%$
$$e=frac{W_o}{W_i}cdot100$$
We can get output work with equation:
$$W_o=mgd_r$$
And we can get input work with equation:
$$W_i=Fd_e$$
Now we get:
$$e=frac{mgd_r}{Fd_e}cdot100rightarrow d_r=frac{eFd_e}{mgcdot100}$$
$$d_r=frac{90cdot250cdot1.6}{150cdot9.81cdot100}$$
$$boxed{d_r=0.24,,rm m}$$
$$e=frac{W_o}{W_i}cdot100$$
We can get output work with equation:
$$W_o=mgd_r$$
Now we have:
$$e=frac{mgd_r}{W_i}rightarrow W_i=frac{mgd_r}{e}$$
$$W_i=frac{215cdot9.81cdot5.65}{72.5}cdot100$$
$$boxed{W_i=16437,,rm J}$$
$$l=2pi r_g$$
$$l=2picdot5$$
$$boxed{l=31.4,,rm cm}$$
$$P_m=Pe_m$$
Now we can get power output of the crane:
$$P_c=P_m e_c=P e_m e_c$$
Now when we have power output of the crane we can calculate velocity with equation:
$$P_c=Fv$$
Where force is weight of the crate, all together we get:
$$v=frac{P e_m e_c}{mg}$$
$$v=frac{5.5cdot10^{3}cdot0.88cdot0.42}{410cdot9.81}$$
$$boxed{v=0.51,,rm m/s}$$
Ideal mechanical advantage of a pulley is given with equation:
$$IMA_p=frac{d_{ep}}{d_{rp}}rightarrow d_{ep}=IMA_pd_{rp}$$
And ideal mechanical advantage of the lever is:
$$IMA_l=frac{d_{el}}{d_{rl}}rightarrow d_{rl}=frac{d_{el}}{IMA_l}$$
In this case $d_{ep}$ is equal to $d_{rl}$, And total ideal mechanical advantage can be write as:
$$IMA=frac{d_{ep}}{d_{rl}}$$
When we put those equations together we get:
$$IMA=frac{IMA_pd_{rp}}{frac{d_{el}}{IMA_l}}=IMA_pcdot IMA_l$$
$$IMA=3cdot2$$
$$boxed{IMA=6}$$
Effectivity is defined as:
$$e=frac{MA}{IMA}cdot100$$
Mechanical advantage can be calculated with:
$$MA=frac{F_r}{F_e}$$
Now we have:
$$e=frac{frac{F_r}{F_e}}{IMA}cdot100rightarrow F_e=frac{F_r}{eIMA}cdot100$$
$$F_e=frac{540}{60cdot6}cdot100$$
$$boxed{F_e=150,,rm N}$$
We can use equation for efficiency:
$$e=frac{W_o}{W_i}cdot100$$
We can get output work:
$$W_o=F_rd_r$$
And we can get input work:
$$W_i=F_ed_e$$
Now we have:
$$e=frac{F_rd_r}{F_ed_e}cdot100rightarrow d_r=frac{F_ed_ee}{F_rcdot100}$$
$$d_r=frac{150cdot12cdot10^{-2}cdot60}{540cdot100}$$
$$boxed{d_r=0.017,,rm m}$$
b) $F_e=150,,rm N$
c) $d_r=0.017,,rm m$
$$W=mgh$$
We can see that there is no difference if she uses longer or shorter ramp.
We can calculate work with equation:
$$W=mgh$$
$$W=240cdot9.81cdot2.35$$
$$boxed{W=5533,,rm J}$$
If a man hold the weights still in place he does not do work.
$$W=0,,rm J$$
We can use the same equation as in part **a** but in this case force is in the opposite direction as velocity and we have to put minus sign.
$$W=-mgh$$
$$W=-240cdot9.81cdot2.35$$
$$boxed{W=-5533,,rm J}$$
A man does not do any work if he let go of the weights.
$$W=0,,rm J$$
We can calculate power with equation:
$$P=frac{W}{t}$$
$$P=frac{5533}{2.35}$$
$$boxed{P=2354,,rm W}$$
b) $W=0,,rm J$
c) $W=-5533,,rm J$
d) $W=0,,rm J$
e) $P=2354,,rm W$
We can calculate the force with equation:
$$cosphi=frac{F_h}{F}rightarrow F=frac{F_h}{cosphi}$$
$$F=frac{805}{cos32^o}$$
$$boxed{F=949,,rm N}$$
We can calculate work with equation:
$$W=F_hd$$
$$W=805cdot22$$
$$boxed{W=17710,,rm J}$$
We can calculate power with equation:
$$P=frac{W}{t}$$
$$P=frac{17710}{8}$$
$$boxed{P=2214,,rm W}$$
b) $W=17710,,rm J$
c) $P=2214,,rm W$
We can calculate work with equation:
$$W=Fd$$
$$W=496cdot2.1$$
$$boxed{W=1042,,rm J}$$
Work done on the refrigerator by machine is equal to difference in potential energy:
$$W_g=mgh$$
$$W_m=115cdot9.81cdot0.85$$
$$boxed{W_m=959,,rm J}$$
Efficiency can be calculated with equation:
$$e=frac{W_o}{W_e}cdot100=frac{W_m}{W}cdot100$$
$$e=frac{959}{1042}cdot100$$
$$boxed{e=92%}$$
b) $W_m=959,,rm J$
c) $e=92%$
We can calculate the force with equation for work:
$$W=Fdcosphirightarrow F=frac{W}{dcosphi}$$
$$F=frac{11.4cdot10^{3}}{25cdotcos48^o}$$
$$boxed{F=681,,rm N}$$
We can calculate force of friction with equation:
$$cosphi=frac{F_f}{F}rightarrow F_f=Fcosphi$$
$$F_f=681cos48^o$$
$$boxed{F_f=456,,rm N}$$
We can calculate the work of the floor with equation:
$$W_f=F_fd$$
$$W_f=456cdot25$$
$$boxed{W_f=11400,,rm J}$$
b) $F_f=456,,rm N$
c) $W_f=11400,,rm J$
$$W=Fdcosphirightarrow phi=cos^{-1}left(frac{W}{Fd}right)$$
$$boxed{phi=58.7^o}$$
To calculate power we can use this equation:
$$P=F_gvsinphi$$
$$P=875cdot0.25cdotsin15^o$$
$$boxed{P=57,,rm W}$$
We can calculate electric power with equation:
$$P_e=frac{P}{e}$$
$$P_e=frac{57}{0.85}$$
$$boxed{P_e=67,,rm W}$$
b) $P_e=67,,rm W$
$$m=frac{150}{30}=5,,rm kg$$
and total number of boxes we have to carry in one go is:
$$n=frac{15}{5}$$
$$boxed{n=3}$$
