Question

A block of mass 1.80 kg is accelerated across a rough surface by a light cord passing over a small pulley as shown in the figure below. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.110 m above the top of the block. The coefficient of kinetic friction is 0.310.

M xーーーーーーーーー

(a) Determine the acceleration of the block when x = .400 m.

(b) Describe the general behavior of the acceleration as the block slides from a location where x is large to x = 0. decreasing, then increasing increasing, or then decreasing no change

(c) Find the maximum value of the acceleration and the position x for which it occurs.

(d) Find the largest value of x for which the acceleration is zero.

Answer 1

The FBD can be seen in the Figure. The law to apply is Newton's second law: ΣΡ-ma The acting forces are: Weight W with magnitude W -mq , m is the mass of the block, g is gravity acceleration, Normal force N with magnitude N , the Tension of cord on block T with constant magnitude T , and Friction force F with magnitude F N μ is coefficient of kinetic friction. Tİh In the horizontal direction, positive direction to the right, from Newton's law, we find: In the vertical direction, positive direction upward: From this equation, we get: N -mg - Tsen0 and, inserting in Tcose - /uN -ma From the geometry (see Figure) tan θ =- → sen θ = x2 +h2 The numerical values are: T= 10.0[N] , m=1.80 [kg]、 9.81 h= 0.110 [m] , μ= 0.310 and g= "(a) Determine the acceleration of the block when x 0.400 10.0 [0.400 0.310 (0.110) 1.80 L V0.4002+0.1102 0.310(9.81) -» a=2.77 Answer to Question (a "(b) Describe the general behavior of the acceleration as the block slides from a location where x is large to r -0: decreasing, then increasing, increasing, or then decreasing, no change " For x large, the behaviour of a is obtained from: lim a- lim - G7μhī)-μg lim a - lim 2 0.310(9.81) a=10.0-0.310(9.81) limo With the values: lim a lim a= 2.52 → - 1.80 80 x-00 10.0 Replacing values: lim a - 0.310 ( -9.81) lim a=-1.32 → 1.80 for large x to a negative value (-1.32 The acceleration goes from a positive value 2.52 small x , then acceleration must take the value zero at some point of this interval for x Respect to the magnitude of the acceleration, we can say: for It diminishes until zero, then increases, when going from large to small values of x :Answer to Question (b)

(c) Find the maximum value of the acceleration and the position x for which it occurs. " The maximum value of the acceleration is achieved when x2 +h2)(2x) dx m x2 +h2 (x2 + h2)3/2-x (x + μ1) which is a transcendental equation in x However, we know from the analysis above, we found that acceleration goes from a maximum positive value for large x to zero, then its magnitude increases again until x-0 , then it is safe to conclude that the maximum acceleration is the value at large x, a -2.52 : Answer to Question (c "(d) Find the largest value ofx for which the acceleration is zero." mg +2x+ (u ih)2-0 , replacing the numerical values: 0.310 (0.110) x + [0.310(0.110)]2 = 0 x2 + 0.097x + 0.001 = 0 → 0.310 (1.80)9.811 10.0 The solution for this equation is: x1 --0.085 [m] and x2 0.012 [m] - x-0.085 [mAnswer to Question (d)