physics question we are all stuck on...?

Question

You and your friends push a 74-kg greased pig up an aluminum slide at the
county fair, starting from the low end of the slide. The coefficient of
kinetic friction between the pig and the slide is 0.070.

(a) All of you pushing together (parallel to the incline) manage to accelerate
the pig from rest at the constant rate of 4.2 m/s2 over a distance of 1.5 m,
at which point you release the pig. The pig continues up the slide, reaching a
maximum vertical height above its release point of 45 cm. What is the angle of
inclination of the slide?
°


(b) At the maximum height the pig turns around and begins to slip down once
slide, how fast is it moving when it arrives at the low end of the slide?
m/s

Answer 1

Lets first find the velocity of the pig when you release the animal

1.5=.5*4.2*t^2
solve for t
t=sqrt(3/4.2)

v=4.2*t
v=sqrt(3*4.2)

v=3.55 m/s
v^2=3*4.2=12.6

Now, the next phase, let's look at energy
KE=.5*74*12.6
When the pig reaches apogee, the KE=0 and the work done by gravity is
74*9.8*.45

and the work done by friction is
74*9.8*.45*0.07/sin(th)

putting it all together

.5*74*12.6=74*9.8*.45+
74*9.8*.45*0.07/sin(th)

I get th=9.4 degrees

To compute the speed at the bottom, the total distance along the slide is 1.5+.45/sin(9.4)
=4.26 meters

The total vertical drop is
.45+1.5*sin(9.4)
=0.695 m
so
.5*m*v^2=m*g*0.695-4.26*m*g*cos(9.4)*0.07


v^2=2*9.8*(0.695-4.26*cos(9.4)*0.07)

v=2.8 m/s

j

Answer 2

I am working the same problem with different numbers... I was able to follow jmurphy's example up to getting the angle, which was 3.6 degrees for mine. However, for the pig sliding down I get that at only 3.6 degrees, friction would prevent the pig from sliding. Not sure what I'm doing wrong... I have the same 74kg pig with .07 friction, theta = 3.59 and I solved for a ramp that is 7.45m long (hypotenuse) and .46m tall. How could you solve this without using energy formulas? Those are next chapter, so this homework problem shouldn't require them.

Answer 3

The rigidity this is pushing the motor vehicle forwards [the engine] has stopped, hence there is no longer something to prepell the motor vehicle forwards on the line except on a hill. the motor vehicle will start to decelerate, or decelerate, till it contains an entire standstill because of friction on the floor [the tyres and tarmac/street floor] and the drag of air. how promptly the motor vehicle stops relies upon at this sort of floor and the fee,

Answer 4

jmurphy has the solution mostly correct.

Vertical displacement is 0.45 not sin(th)*0.45
work done by gravity is
74*9.8*0.45

and the work done by friction is
0.07*74*9.8*0.45/tan(th)

putting it all together
.5*74*3*4.2 = 74*9.8*0.45 + 0.07*74*9.8*0.45/tan(th)
139.86 = 50.764/tan(th)
theta = Tan^-1 (50.764/139.86)
theta = 19.95 degrees