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Hc verma +1 Ch-8 Pg. 134 Ques no. 28 and 31 Please

Hc verma +1 Ch-8 Pg. 134 Ques no. 28 and 31 Please

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1 Answers

Amogh Gajare
60 Points
6 years ago
H.C. Verma Ch. 8 Problem 28.
m = 30 kg.
v = 40 cm/s = 0.4 m/s
s = 2m
By the free body diagram,
F = m(a -g)
also, a = \frac{v^2 - u^2}{2s}
 
a = \frac{0.16}{0.4} = 0.04 m/s
 
Work done = W = F.s.cos\theta
W = m(a - g)*s*cos\theta
After calculation we get
W = – 586J
 
Problem 31 
 
m= 4 kg
v2 = 0.3 m/s
v1 = 2 * 0.3 = 0.6 m/s
h = 1 m
s = 2*1 = 2 m
u = 0
Therefore,by work energy theorem,
[\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2] - 0 = - \mu Rs + m_2g...................R = 40 Newtons
On calculation,
\mu = \frac{9.235}{80} = 0.12
 

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