A 3.3 kg block, initially in motion, is pushed along ahorizontal floor by a forc

Question

A 3.3 kg block, initially in motion, is pushed along ahorizontal floor by a force F of magnitude 17 N at an angle 45degrees with the horizontal. The coefficient of friction betweenthe block and the floor is .25. a) calculate the magnitude of the frictional force on theblock from the floor b) calculate the blocks acceleration A 3.3 kg block, initially in motion, is pushed along ahorizontal floor by a force F of magnitude 17 N at an angle 45degrees with the horizontal. The coefficient of friction betweenthe block and the floor is .25. a) calculate the magnitude of the frictional force on theblock from the floor b) calculate the blocks acceleration

Explanation / Answer

Data:        m = 3.3kg    F =17N    = 450 =0.25 We know that F cos 45 >> moves forward F sin 45 >> upwards, counters weight Normal force = N = mg - F sin 45 Friction force (f) = u * N = 0.25*[3.3*9.8 - 17*0.707] f = 5.080 N Then net force (forward) = ma = Fcos 45 - f ma = 6.940N Then F = ma       a = F/m          = 6.940/3.3 a = 2.103 m/s2 = acceleration of block

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