Question #37265

The drawing shows two boxes resting on frictionless ramps. One box is relatively light and sits on a steep ramp. The other box is heavier and rests on a ramp that is less steep. The boxes are released from rest at A and allowed to slide down the ramps. The two boxes have masses of 9 and 48 kg. If A and B are 5.5 and 0.5 m, respectively, above the ground, determine the speed of (a) the lighter box and (b) the heavier box when each reaches B. (c) What is the ratio of the kinetic energy of the heavier box to that of the lighter box at B?
http://edugen.wileyplus.com/edugen/courses/crs3976/art/qb/qu/c06/qu_6_38.gif

Expert's answer

SOLUTION

from the conservation energy law : reduction of potential energy equals to increase of kinetic energy

for a small box

Ep=m*g*h - reduction of the potential energy

g=10H/kg

h=(A-B)=5m

Ep=9*10*5=450J

Ek=m*V^2/2 - increase of the kinetic energy

so Ep=Ek

m*g*h=m*V^2/2

g*h=V^2/2

V^2=2*g*h

V=sqrt(2*g*h)=sqrt(2*10*5)=sqrt(100)=10

V=10m/s - velocity does NOT depend on mass

so

a) velocity of the lighter box V_light=10m/s

b) velocity of the heavier box V_heavy=10m/s

c) ratio of the kinetic energy

m_heavy=48kg

m_light=9kg

R=Ek_heavy/Ek_light=(m_heavy*V^2/2) / (m_light*V^2/2)=m_heavy / m_light = 48/9= 5.33

R=5.33

from the conservation energy law : reduction of potential energy equals to increase of kinetic energy

for a small box

Ep=m*g*h - reduction of the potential energy

g=10H/kg

h=(A-B)=5m

Ep=9*10*5=450J

Ek=m*V^2/2 - increase of the kinetic energy

so Ep=Ek

m*g*h=m*V^2/2

g*h=V^2/2

V^2=2*g*h

V=sqrt(2*g*h)=sqrt(2*10*5)=sqrt(100)=10

V=10m/s - velocity does NOT depend on mass

so

a) velocity of the lighter box V_light=10m/s

b) velocity of the heavier box V_heavy=10m/s

c) ratio of the kinetic energy

m_heavy=48kg

m_light=9kg

R=Ek_heavy/Ek_light=(m_heavy*V^2/2) / (m_light*V^2/2)=m_heavy / m_light = 48/9= 5.33

R=5.33

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