A crate of mass 36kg is released from the top of a ramp. The ramp is 2.5m lang and inclined at an angle of 0 (theta) to the horizontal, where sin 0 (theta) =3/5. As the crate reaches the bottom of the ramp, it gains a speed of 4.5m/s.
(a) Calculate the gain in kinetic energy.
(b) Calculate the loss in potential energy.
(c) Compare the results in (a) and (b), and show that some other form/s of energy must have been produced.
(d) What is the nature and size of the other form/s of energy in (c)?
(e) Calculate the friction between the crate and the ramp.
(f) Find the work done by gravitational force.
2007-09-10 13:28:52 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The loss of potential energy is
36*10*2.5*3/5
540 J
At the bottom of the ramp the crate has kinetic energy of
.5*36*4.5^2
364.5 J
The difference is
175.5 J, which is lost to friction (heat)
Assuming all friction from the ramp crate interface, the force is
f*2.5=175.5
f=70.2 N
The work done by gravity is the loss in potential energy 540 J
j
2007-09-10 22:07:09 · answer #1 · answered by odu83 7 · 0⤊ 0⤋