Asked by gy
In the figure, a 2.4 kg box of running shoes slides on a horizontal frictionless table and collides with a 1.4 kg box of ballet slippers initially at rest on the edge of the table, at height h = 0.75 m. The speed of the 2.4 kg box is 4.6 m/s just before the collision. If the two boxes stick together because of packing tape on their sides, what is their kinetic energy just before they strike the floor?
Answers
Answered by
Elena
m₁=2.4 kg, m₂=1.4 kg, v₁=4.6 m/s, v₂=0
m₁v₁= (m₁+m₂) u
u= m₁v₁/(m₁+m₂) =2.4•4.6/(2.4+1.4) =2.9 m/s
At the edge of the table, two boxes have
kinetic (KE) and potential energy (PE), and near the ground - KE₁
KE + PE=KE₁
(m₁+m₂)u²/2 + (m₁+m₂)gh =(m₁+m₂)v²/2
u²/2 +gh = v²/2
v=sqrt{ u²+2gh} =sqrt{2.9² +2•9.8•0.75}=3.65 m/s
m₁v₁= (m₁+m₂) u
u= m₁v₁/(m₁+m₂) =2.4•4.6/(2.4+1.4) =2.9 m/s
At the edge of the table, two boxes have
kinetic (KE) and potential energy (PE), and near the ground - KE₁
KE + PE=KE₁
(m₁+m₂)u²/2 + (m₁+m₂)gh =(m₁+m₂)v²/2
u²/2 +gh = v²/2
v=sqrt{ u²+2gh} =sqrt{2.9² +2•9.8•0.75}=3.65 m/s
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