A 3.3 kg block, initially in motion, is pushed along a horizontal floor by a for

Question

A 3.3 kg block, initially in motion, is pushed along a horizontal floor by a force F of magnitude 17 N at an angle ? = 50 with the horizontal (Figure 6-20). The coefficient of kinetic friction between the block and floor is 0.25. (Assume the positive direction is to the right.) (a) Calculate the magnitude of the frictional force on the block from the floor. __________N (b) Calculate the magnitude of the block's acceleration. ________m/s2 Preferences 1 2 3 4 5 6 7 8 9 0 - = Backspace Tab q w e r t y u i o p [ ] Return capslock a s d f g h j k l ; ' shift ` z x c v b n m , . / shift English Deutsch Espaol Franais Italiano Portugus ??????? alt alt Preferences

Explanation / Answer

taking force components along vertical direction and balancing them

N = mg + F sin(theta)

Friction force = uk* N = .25 * (3.3*9.81 + 17*sin(50)) = 11.35 N

this friction force is backwards that is in negative direction

b)Consider forces along horizontal direction

Fcos(theta) - friction = ma

17cos(50) - 11.35 = 3.3*a

solving this gives acceleration a in negative.

a = -0.128 m/s^2

Which means block is decelerating . with the applied force, friction force has increased and the block is slowing down
Acceleration magnitude = 0.128

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