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March 27, 2015

March 27, 2015

Posted by **R** on Wednesday, May 8, 2013 at 12:01am.

rB = 10 m,

rC = 20 m,

The heights at points A, B, and C are

hA = 22 m,

hB = 32 m,

and

hC = 0.

Assume friction is negligible and ignore the kinetic energy of the wheels. (The figure is not necessarily drawn to scale.)

What is the weight at B?

- Physics -
**Elena**, Wednesday, May 8, 2013 at 6:10amPoint A: h₁= 22 m, v₁=16 m/s;

Point B: h₂= 32 m, r₂= 10 m, v₂=?;

Δh₁₂ = h₂-h₁=32-22 = 10 m;

Point C: h₃=0, r₃= 20 m, v₃=?;

Δh₂₃ = h₂-h₃ = 32 – 0 = 32 m.

KE₁+PE₁=PE₂ + KE₂,

mv₁²/2 +mgh₁=mgh₂ +mv₂²/2,

mv₁²/2= mv₂²/2+ mgh₂ - mgh₁=

=mv₂²/2+ mg Δh₁₂.

v₂ = sqrt{ v₁²- 2gΔh₁₂} =

=sqrt{16² - 2•9.8•10} =7.75 m/s.

For point B:

ma₂=mg-N₂,

N₂=mg-ma₂=m[g- (v₂²/r₂)]=

= 26[ 9.8 – (7.75²/10)]= 98.8 N,

N₂ (normal force) = W₂ (weight)

W₂= 98.8 N.

PE₂ +KE₂ = PE₃+KE₃,

mgh₂ +mv₂²/2 = mgh₃ +mv₃²/2,

mv₃²/2 = mgh₂ +mv₂²/2 - mgh₃=

=mv₂²/2 + mg Δh₂₃ ,

v₃ = sqrt{v₂² +2g Δh₂₃}=

=sqrt{ 7.75² + 2•9.8•32) = 26.2 m/s.

For point C:

ma₃= N₃ – mg,

N₃= mg+ma₃=m[g+ (v₃²/r₃)]=

= 26[ 9.8 + (26.2²/20)]= 1148.2 N.

N₃ (normal force) = W₃ (weight)

W₃= 1148.2 N.

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