A block of mass 2 00 kg is accelerated across rough surface by a ight cord pesar

Question

A block of mass 2 00 kg is accelerated across rough surface by a ight cord pesar over s ul pulley as shown r1 the figure below. The tens on rl the co d is main ained ut 10.0 N und he pulley is 0.120 r esove the top u he block. The ooerficient u kinetic ricton 0.330 a) Determine the aceleration of the blode when x =·nou m. (b) Desoribe the general behavio of the acoeleration as the block slides from a location wherexis large tox-0 increasing, then decreusing O decreesiry, then increasing no change (c Find the maximurn value of the acceleration and the position x for which it occurs. m/S2 (d) Find the largest valua af x far which the acceleration is zera. Need Help?

Explanation / Answer

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

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