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H.C. Verma book
page 135
Q. no........41,43
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
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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