Work, Energy and Power

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New answer posted•a year ago
S
a year ago

a = k2rt2

⇒ v 2 r = k 2 r t 2

⇒ v = k r t

a t = d v d t = k r

? tangential force, Ft = mat = mkr

∴ P o w e r     d e l i v e r e d , P = F t v = ( m k r ) ( k r t ) = m k 2 r 2 t  

 Note ® Power delivered by centripetal force will be zero.

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New answer posted•a year ago
S
a year ago

According to conservation of energy, we can write

Gain in kinetic energy = Loss in potential energy

⇒  Kf – Kin = Uin - Uf

⇒ K - = mgy – mg (y – y0) = mgy0

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

ΔW = area under P — t graph
= (1/2) (4 + 6) × 7 = 35 J
Work done = change in KE ⇒ 35 = (1/2) × 2 × v² - (1/2) × 2 × (1)² ⇒ v = 6 m/s

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Vertical component of velocity just after collision = u 2 2

k i = m u 2 2 k f = 1 2 m u 2 2 + 1 2 m u 2 8 = 5 m u 2 16

Fraction  = k i - k f k i = 1 - 5 m u 2 16 m u 2 2 = 1 - 5 8 = 3 8

 

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New answer posted•a year ago
S
a year ago

To reach at point A particle must cross the peak point.

Loss in KE = gain in PE

1 2 × 1 0 0 1 0 0 0 × v 2 = ( 5 − 0 )            

v2 = 100

v = 10 m/s

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New answer posted•a year ago
S
a year ago

  1 2 μ v r 2 = 1 2 k x 2  

  x = V r e r 2 μ k          

= 2 × 8 / 6 1 0 = 8 1 5            

From conservation of momentum

2 × 4 + 4 × 2 = 2v1 + 4v2

8 = v1 + 2v2                     

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New answer posted•a year ago
S
a year ago

When relative sliding stops, both move with same velocity.
Only friction acts in horizontal direction.
∴ for system no external force, we can apply conservation of momentum.
1 × 12 + 5 × 0 = 6 × v
v = 2 m/s
Now work done by friction on block = change in kinetic energy of block
w = (1/2)m (2² - 12²) = (1/2

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New answer posted•a year ago