A block of mass 2.20 kg is accelerated across a rough surface by a light cord pa

Question

A block of mass 2.20 kg is accelerated across a rough surface by a light cord passing over a small pulley as shown in Figure P5.73. The tension T in the cord is maintained at 10.0 N, and the pulley is 0.100 m above the top of the block. The coefficient of kinetic friction is 0.400. (a) Determine the acceleration of the block when x = 0.400 m. (b) Describe the general behavior of the acceleration as the block slides from a location where .vis large to x = 0. (c) Find the maximum value of the acceleration and the position .x for which it occurs, (d) Find the value ol .v for which the acceleration is zero.

Explanation / Answer

Part. A. Find the angle for the specific case when x=0.4m. The angle should be =arctan(0.1/0.4) = 14.04degrees. Part. B. Sum forces in x and y direction. For the sum in the "y"direction, you should find that the forces sum to zero like: 0 = T*sin + N - M*g Forces in "x" direction: M*a = T*cos - *N You can use the sum in the "y" direction to solve for thenormal force N, then plug into the "x" direction force balance tosolve for a. a = (1/M)*(T*cos - *(M*g - T*sin ). Part C. You can find the derivative of the accelerationwith respect to theta, which is indirectly associated with x. If you take the derivative of the expression above and set it equalto zero, you'll find that the tan = y/x = a numericalvalue. Then x = y/numerical value. Evaluate what theangle is and plug in to the formula above to find the value of maxacceleration. Part D. Set a=0, solve for theta. Once you knowthe angle, you can set cos =x/(x^2+y^2). Solvefor x and you'll have the position of zero acceleration. Part. A. Find the angle for the specific case when x=0.4m. The angle should be =arctan(0.1/0.4) = 14.04degrees. Part. B. Sum forces in x and y direction. For the sum in the "y"direction, you should find that the forces sum to zero like: 0 = T*sin + N - M*g Forces in "x" direction: M*a = T*cos - *N You can use the sum in the "y" direction to solve for thenormal force N, then plug into the "x" direction force balance tosolve for a. a = (1/M)*(T*cos - *(M*g - T*sin ). Part C. You can find the derivative of the accelerationwith respect to theta, which is indirectly associated with x. If you take the derivative of the expression above and set it equalto zero, you'll find that the tan = y/x = a numericalvalue. Then x = y/numerical value. Evaluate what theangle is and plug in to the formula above to find the value of maxacceleration. Part D. Set a=0, solve for theta. Once you knowthe angle, you can set cos =x/(x^2+y^2). Solvefor x and you'll have the position of zero acceleration.

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