a crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100N parallel to the incline, which makes an angle of 20.0 degrees with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m. (1) How much work is done by gravity? (b) how much mechanical energy is lost due to friction? (c) what is the change in kinetic energy of the crate? (e) what is the speed of the crate after being pulled 5.00 m?
2007-10-28 09:41:07 · 2 answers · asked by Mika 2 in Science & Mathematics ➔ Physics
1) Gravity will be negative as
-10*9.81*5*sin(20)
=-168 J
2) Friction
5*10*9.81*0.4*cos(20)
=184 J
3) Kinetic energy
The energy equation is
(KEo-KEi)=100*5-168-184
=148 J
4) .5*10*v^2=.5*10*1.5^2+148
v=sqrt(1.5^2+2*148/10)
=5.64 m/s
j
2007-10-28 10:41:59 · answer #1 · answered by odu83 7 · 0⤊ 0⤋
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2016-10-14 06:53:35 · answer #2 · answered by kosmoski 4 · 0⤊ 2⤋