As shown below, a block of mass 5.00 kg is placed on top of a larger block of ma

Question

As shown below, a block of mass 5.00 kg is placed on top of a larger block of mass 14.0 kg, and the two-block system rests on a horizontal surface. There is friction between the two blocks, as well as between the larger block and the surface, with static and kinetic friction coefficients of S = 0.300 and K = 0.200, respectively. As shown below, a horizontal force F is applied to the larger block. Use g = 10 m/s2, and try doing this problem without a calculator.

(a) If the blocks are initially at rest, what is the minimum value of the horizontal force F necessary to start the system moving?

(b) Once the blocks are moving, what is the minimum value of the horizontal force F necessary to keep the system moving at constant velocity?

(c) Once the blocks are moving, what is the maximum value the horizontal force F can be such that the small block does not slide on the large block?

Explanation / Answer

Total normal force by ground on large block = (5+14)*10 = 190 N

Maximum state friction = N*S = 190*0.3 = 57 N

a) So minimum value of the horizontal force F necessary to start the system moving = 57 N

b) dynamic friction = 190 * 0.2 = 38 N

So Once the blocks are moving, the minimum value of the horizontal force F necessary to keep the system moving at constant velocity = 38 N

c) mimimum force required on top block for it to move = 5*10*0.3 = 15 N

To have 15 N force the acceleration of the block system should be 15/5 = 3 m/s2

Force reuired to create 3 m/s2 acceleration = frictional force (dynamic) + (m+M)*3 = 38+19*3 = 95 N

So maximum value the horizontal force F can be such that the small block does not slide on the large block = 95 N

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