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# H.C. Verma bookpage 135Q. no........41,43

## 2 Answers

8 years ago

41. m = 5kg,    x = 10cm = 0.1m,    v = 2m/sec,

h =? G = 10m/sec2

So, k =mg/x

50/0.1 = 500 N/m

Total energy just after the blow E = ½ mv2 + ½ kx2 …(i)

Total energy at a height h = ½ k (h – x)2 + mgh …(ii)

½ mv2 + ½ kx2 = ½ k (h – x)2 + mgh

On solving we can get,H = 0.2 m = 20 cm

43. m = 2kg, s1 = 4.8m,

x = 20cm = 0.2m,

s2 = 1m,  sin 37° = 0.60 = 3/5,  Ø = 37°,  cos 37° = .79 = 0.8 = 4/5

g = 10m/sec2

Applying work – Energy principle for downward motion of the body

0 – 0 = mg sin 37° × 5 – μR × 5 – ½ kx2

»20 × (0.60) × 1 – μ× 20 × (0.80) × 1 + ½ k (0.2)2 = 0

»60 – 80μ - 0.02k = 0

» 80μ + 0.02k = 60 …(i)

Similarly, for the upward motion of the body the equation is

0 – 0 = (–mg sin 37°) × 1 – μ R × 1 + ½ k (0.2)2

» –20 × (0.60) × 1 – μ ×20 × (0.80) × 1 + ½ k (0.2)2 = 0

» –12 – 16μ + 0.02 K = 0 ..(ii)

Adding equation (i) & equation (ii), we get 96 μ = 48

» μ = 0.5

Now putting the value of μ in equation (i) K = 1000N/m

7 years ago

What is the work – Energy principle for downward motion of the body?

and work – Energy principle for upward motion of the body?

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