A block of weight 120 N is at rest on a horizontal surface A horizontal force of

Question

A block of weight 120 N is at rest on a horizontal surface A horizontal force of 54 N is required to set the block in motion and a horizontal force of 42 N is required to keep it moving with a constant speed. The coefficient of kinetic friction is 0.45. 0.54. 0.42. 0.35. A constant force F = (4.0 i - 3.0 j) N acting on a block pushes the block through a horizontal displacement s = 0.5 (m) i. The work done by the force is 1.5 J. 2.0 J. 2.5 J. -1.5 J. A block of mass 0.20 kg with a speed 3.0 m/s collides with another block of the same mass at rest on a frictionless horizontal surface. The kinetic energy of the first block before the collision is 0.6 J. 0.9 J. 0.3 J.

Explanation / Answer

10) m = w/9.8 = 120/9.8 = 12.24 kg

ma max = mues*mg

a max = mues*g

mues = amax/g = (54/12.24)/9.8 = 0.45

19) w = f.s = 4*0.5(i.i)-3*0.5(j.i)

i.i = 1

j.i = 0

w = 2 J

20) ki = 1/2mv^2 = 1/2*0.2*3^2 = 0.9 J

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