Ive been stuck on these problems.
1) A crate slides down an inclined plane without friction. If it is released from rest and reaches a speed of 5.649 m/s after sliding a distance of 2.24 m, what is the angle of inclination of the plane with respect to the horizontal?
2)An archer shoots an arrow from a height of 1.14 m above ground with an initial velocity of 39.0 m/s and an initial angle of 39.8° above the horizontal. At what time after the release of the arrow from the bow will the arrow be flying exactly horizontally?
3)A football is punted with an initial velocity of 22.5 m/s and an initial angle of 57.5°. What is its hang time (the time until it hits the ground again)?
4)An elevator cabin has a mass of 368.5 kg, and the combined mass of the people inside the cabin is 178.4 kg. The cabin is pulled upward by a cable, with a constant acceleration of 3.22 m/s2. What is the tension in the cable?
5)A store sign of mass 9.95 kg is hung by two wires that each make an angle of θ = 44.0° with the ceiling. What is the tension in each wire?
Answers & Comments
1) we can calculate the acceleration using V^2=V(0)^2+2*a*d
5.649^2=0^2+2*a*2.24
a=7.1
If you draw a force graph you will see that the force that is causing this acceleration equals to F(g)*sin(alpha) , when alpha is the angle of the slope with the horizontal. so:
m*a = m*g*sin(alpha)
sin(alpha) = (ma)/(mg) = a/g = 7.1/9.8 = 0.72
alpha = 46.4 degrees
2) the arrow will fly horizontally at the maximum height, because at this point it's vertical speed becomes zero and so it only has horizontal speed.
the initial vertical speed equals to 39*sin(39.8) or 25m/s.
now we can calculate how long it takes for the vertical speed to become 0:
V=V(0)+a*t
0=25+(-9.8)*t
t=2.55
3) same as last question, the horizontal speed is not relevant but only the vertical speed.
only this time instead of how long it takes to get to maximum height we need the time it takes to come back down again. in this case when it comes back down it will have the same speed as the initial vertical speed only downwards. so if the initial speed is 22.5*sin(57.5)=19m/s then the touchdown speed will be -19m/s.
V=V(0)+a*t
-19=19+(-9.8)*t
t=3.9
4) the force the cable exerts on the elevator is
F=ma=(368.5+178.4)*3.22=1761
5) since the sign is not falling, that means that the vertical portion of the force both wires exerts equals to the force of gravity of the sign (again draw a force graph). so, assuming the load is spread evenly on both wires, the vertical portion of the force that a single wire exerts equals to half the force of gravity on the sign or 0.5*m*g or 0.5*9.95*9.8 = 48.755N.
so that is the vertical portion of the force the wire exerts. then the force itself would be 48.755/sin(44) or 69.65N.