Two forces, FA and FB, act on an object such than the object accelerates in the

Question

Two forces, FA and FB, act on an object such than the object accelerates in the y, direction as shown in figure. If FA is 175 N,(a) What is the magnitude of FB? (b) What is the acceleration of the object if its mass is 50.0 kg? A 15.0-kg block slides down along a 30degree incline from a height of 25.0m. (a) If the block roaches the bottom of the slope at 15.0 m/s, what is the average factional force between the block and the surface of the slope? (b) What is the kinetic coefficient of friction of the block on the slope? (c) After reaching the bottom of the slope, the block continues to slide on the level ground for 25.0 m before slopping, what is the kinetic coefficient of friction between the level ground and the block?

Explanation / Answer

14

a) FA * sin 45 = FB * sin 30

175 * sin 45 = FB sin 30

so. FB = 247.4874 N

b) total force on object = FA cos 45 + FB cos 30 = 175*sin 30 + 247.4874*cos 30 = 365.8848 N

so... accelelarion = force / mass = 365.8848/50 = 7.3177 m/sec2

15

a) initial potential energy = 15*9.8*25 = 3675 J

final energy = 0.5*15*15^2 =1687.5 J

so... work done by friction force = inital enerygy - final = 3675 - 1687.5 = 1987.5 J

length along the incline = height / sin 30 = 25 / sin 30 = 50 m

so... Friction force * 50 = 1987.5

so F = 1987.5/50 = 39.75 N

b)

normal reaction by incline = mg*cos 30 = 15*9.8*cos 30 = 127.3057 N

so.. coefficient of friction factor= 39.75/127.3057 = 0.31224

c) energy of block after reaching ground = 1687.5 J

so.. work done by friction = 1687.5 J

normal reaction by ground = m*g = 15*9.8 = 147 N

length tavelled = 25 m

..

so.. friction * 25 = 1687.5

so.. friction = 1687.5 /25 = 67.5 N

so.. coefficient of frcitoin = 67.5/147 = 0.45918

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