Question 662387
1. The distance (in km) traveled by a car moving at constant 40 km/hr for 3 hours is
{{{3*40=120}}}.
The distance (in km) traveled by a car moving at constant 40 km/hr for 2 hours is
{{{2*40=80}}}.
If the car was traveling {{{120}}} km east for the first 3 hours,
and then traveled {{{80}}} km north for the next 2 hours,
we can represent the trip like this:
{{{drawing(300,300,-1.5,13.5,-2.5,12.5,
triangle(0,0,12,0,12,8),
arrow(3,4,3,10),locate(2.8,11,N),
arrow(0,7,6,7),locate(6.2,7.3,E),
locate(-1,-0.2,start),locate(11.5,8.8,end),
locate(5.7,-0.1,120km),locate(12.1,4.5,80km),
locate(6,4,d),rectangle(12,0,11.7,0.3)
)}}} The distance, from start to end, in km, is {{{d=sqrt(120^2+80^2)}}}=about{{{highlight(144)}}} km.
 
2. A {{{20}}} ft ladder, leaning against {{{25}}} ft vertical wall, making a {{{75^o}}} angle with horizontal ground, can be represented as
{{{drawing(150,300,-3,7,-3,27,
line(-3,0,7,0), line(0,0,0,25),
blue(line(5.18,0,0,19.3)),
rectangle(0,0,0.5,0.5),
locate(3,1.8,75),locate(4,2.2,o),
locate(2.6,10.5,20ft),locate(-2.9,13.2,25ft),
arrow(-0.8,8.7,-0.8,0),arrow(-0.8,10.5,-0.8,19.32),
arrow(-2,11.6,-2,0),arrow(-2,13.4,-2,25),
locate(-1,10.3,h)
)}}} In ft, {{{h=20*sin(75^o)}}}= about{{{20*0.966=19.32}}} (rounding)
How far are the tips of the ladder from the top edge of the wall?
In feet, {{{25-19.32=5.68}}}. I would say 5 ft 8 inches.
 
3. A boy standing in front of a cliff claps his hands, and 2.50 seconds later he hears an echo. How far away is the cliff if sounds travel at 343 m/sec?
The sound traveled a distance {{{d}}} to the cliff and a distance {{{d}}} back to the boy
in {{{2.50s}}} at {{{343m}}}/{{{s}}}
{{{2d=2.50*343m}}} --> {{{d=2.50*343/2}}}{{{m}}} --> {{{highlight(d=429m)}}} (rounded to 3 significant figures)
 
4. A motorbike has an initial velocity of {{{20m}}}/{{{s}}} and accelerates at -1.0{{{m}}}/{{{s^2}}}.
NOTE: I entered reasonable units, and assume the acceleration is really negative; the motorcycle is slowing down).
4a Finds its velocity after 10 seconds {{{v(t)=v[0]+a*t}}} 
{{{v(0)=20+(-1.0)*10}}} --> {{{v(10)=10m}}}/{{{s}}}
4b What its displacement after 10 seconds
We could multiply {{{10s))) times the average velocity {{{15m}}}/{{{s}}} over those 10 seconds, to get {{{displacement=150m}}}
We could use {{{d=v[0]*t+(1/2)a*t^2}}} --> {{{d=20*10+(1/2)(-1)*10^2}}} --> {{{d=200+(1/2)(-1)*100}}} --> {{{d=200-50}}} --> {{{highlight(d=150m)}}}
It is 150 m further ahead in the direction it was originally moving.
4c give its velocity after 30 seconds
Very funny!
If the motorcycle keeps accelerating at -1.0{{{m}}}/{{{s^2}}}, it will stop after 20 seconds and will start moving in the opposite direction at increasing speed.
I hope the bike rider just made a U-turn, and is not riding backwards,
In that case, the final velocity will be {{{-10}}} m/s.
That is {{{36000m}}} in one hour, {{{36kph}}}.
 
5. A 10 kg block of wood is at rest on a level concrete floor.
Taking the acceleration of gravity as {{{g=9.8m}}}/{{{s^2}}},
the normal (perpendicular) force pressing the block to the concrete (the block's weight) is {{{98N}}}.
5a What will happen if horizontal force of 50N is applied to the block?
I would have thought nothing, because the static friction coefficient between wood and concrete is often quoted as 0.62, but part b suggest that your concrete is more slippery than that. Maybe it is, and the block will start moving.
5b What is the minimum horizontal force needed to start the block moving if coefficient of static friction is = 0.5?
That would be {{{0.5*98N=highlight(49N)}}}
5C What is the minimum horizontal force needed to keep the block in motion at constant speed if coefficient of kinetic friction is =0.4?
That would be {{{0.4*98N=39.2N}}}.
I'd answer {{{highlight(39N)}}} because all the given data had just 1 or 2 significant digits.
 
6. A 70 kg man is fired from a compressed air canon and emerged from the 20 m barrel at 40 m/s, Using Newton's Second Law, find the average force on him during the firing of the cannon.(Start by finding the acceleration with the Formula {{{V^2 =V[0]^2 +2 ax}}} and then {{{F=ma}}}).
The final velocity, on exiting the barrel is {{{V=40}}} m/s.
The initial velocity, before being accelerated, is {{{V[0]=0}}} m/s, of course.
The distance covered while accelerating is the length of the barrel, so {{{x=20}}} m.
Substituting in the formula above,
{{{40^2=0^2+2*a*20}}} --> {{{40^2=40a}}} --> {{{highlight(a=40)}}} {{{m}}}/{{{s^2}}}
Then {{{F=ma}}}, with {{{m=70}}} kg, results in
{{{F=70*40}}} N --> {{{highlight(F=280)}}} N
 
7. Compare the potential energy of a 1,500 kg car of the top at a hill 50 m high with its kinetic energy when moving at 100 km/hr?
Using {{{g=9.8m}}}/{{{s^2}}}, and {{{U=mgh}}} we calculate the potential energy of the car at the top the hill as
{{{U=mgh}}} --> {{{U=1500*9.8*50}}} --> {{{highlight(U=735000)}}} J.
The kinetic energy can be calculated as {{{K=(1/2)mv}}},
and with {{{100km=100000m}}} and {{{1hour=3660s}}}, {{{v=100000/3600=27.78}}} m/s,
so for the 1500 kg car moving at 100 km/hr
{{{K=(1/2)*1500*27.78^2}}} --> {{{K=(1/2)*1500*771.6}}} --> {{{highlight(K=579000)}}} J.
The potential energy was greater tan the kinetic energy.
If the potential energy could efficiently convert to kinetic energy, the car would have more than enough energy to reach 100 km/hr by just rolling down the hill.
 
8.A 4.5 kg rifle fires a 10 g bullet with a velocity of 1000 m/sec. What is the recoil velocity of the riffle?
The mass of the 10 g bullet, in kg is 0.010 kg.
Conservation of momentum requires that the momentum (mass times velocity) of the riffle must be the same in magnitude (but with opposite direction) as the momentum of the bullet, so the recoil velocity, {{{V}}} , in m/s satsifies the equation
{{{4.5*v=0.01*1000}}} --> {{{4.5v=10}}} --> {{{v=10/4.5}}} --> {{{highlight(v=2.2)}}} m/s (rounded).