An object slides down an inclined plane of angle 30° and of incline length 5 m. If the initial speed of the object is 5 m/s directed down the incline, what is the speed at the bottom? Neglect friction.
2007-02-28 12:23:14 · 3 answers · asked by Mike B 1 in Science & Mathematics ➔ Physics
kinetic energy at bottom is the sum of the potential and the kinetic energy at the top.
2007-02-28 18:46:16 · answer #1 · answered by meg 7 · 0⤊ 0⤋
Using the conservation of energy principle, the energy at the top of the incline is equal to the energy at the bottom, i.e.:
Initial Kinetic Energy (KE) + Initial Potential Energy (PE)=
Final KE+Final PE.
KE=1/2mv2
where m is the mass in kg, and v the speed in m/s.
PE=mgh
where m is the mass in kg, g the acceleration of gravity equal to 9.8m/s^2, and h is the displacement in meters.
Substitute known values:
1/2m*5^2+mg*5*sine30=
1/2mv^2+mg*0
m cancels out.
12.5+9.8*5*0.5=1/2v^2
37=1/2v^2
v^2=37*2
v^2=74
v=8.6m/s
2007-03-01 02:54:52 · answer #2 · answered by tul b 3 · 0⤊ 0⤋
it fairly is going to look confusing at first yet attempt utilising the freebody diagram. because of the fact the physique is on an vulnerable airplane, the conventional reaction R = mgcos30 ( the two will stability one yet another) F = mgsin30 = ma. if the incline is tender, then the above family will prepare. if the incline is tough , then a coefficient of friction would be present day (?), then the equation would be mgsin30 - ?mgcos30 = ma. a = (gsin30 - ?cos30) if the preliminary velocity is u , then the fee on the backside (v) would be v^2 - u^2 = 2as
2016-09-30 01:01:52 · answer #3 · answered by ? 4 · 0⤊ 0⤋