given, m = 2kg
T = 10 N
h = 0.12 m
k = 0.33
a. when x = 0.4 m
Normal reaction = N
N + T*h/sqroot(h^2 + x^2) = mg
N = 16.7465 N
hence
T*x/sqroot(h^2 + x^2) - kN = m*a
a = 2.025955 m/s/s
b. N + T*h/sqroot(h^2 + x^2) = mg
dN/dx = T*hx/(h^2 + x^2)^3/2
T*x/sqroot(h^2 + x^2) - kN = m*a
T(x^2/sqroot(h^2 + x^2) - sqroot(h^2 + x^2))/(h^2 + x^2) - kT*hx/(h^2 + x^2)^3/2 = m*da/dx
m*da/dx = T(-h^2))/(h^2 + x^2)^3/2 - kT*hx/(h^2 + x^2)^3/2
hence da/dx is -ve
hence
a decreases as x increases
c. for maximum value of acceleration
da/dx = 0
T(-h^2))/(h^2 + x^2)^3/2 - kT*hx/(h^2 + x^2)^3/2 = 0
T(-h^2)) - kT*hx = 0
h = -kx
x = -h/k = -0.363 m
a max = -7.46755 m/s/s
d. for a = 0
N = T*x/sqroot(h^2 + x^2)k
(xk + h)^2 = (mg/T)^2*(h^2 + x^2)
0.1089x^2 + 0.0144 + 0.0792x = 3.849444(0.0144 + x^2)
3.740544x^2 - 0.0792x + 0.0410319936 = 0
acceleration of the block is never 0
A block of mass 2 00 kg is accelerated across α rough surface by a ight...
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