If anyone can help it\'ll be greatly appreciated. The coefficient of friction be

Question

If anyone can help it'll be greatly appreciated.

The coefficient of friction between the 6.9 kg mass and the table is 0.55, and the coefficient of friction between the 5 kg mass and the table is 0.31. Consider the motion of the 33 kg mass which descends by the amount of Delta y = 0.42 m after releasing the system from rest. The acceleration of gravity is 9.8 m/s2 . Find the work done against friction. Answer in units of J Find the speed of the 33 kg mass. Answer in units of m/s Find the time it takes for the 33 kg mass to fall a distance of Delta y = 0.42 m.

Explanation / Answer

work done against friction = -(0.55*6.9g + 0.31*5g)*0.42 = -22J

33g-T13 = 33a

T13 - T12 - 0.31*5g = 5a

T12 - 0.55*6.9g = 6.9a

adding we get

a = 6.036 m/s2

so now v2 - u2 = 2as

where u = 0, s = 0.42m and a=6.036

gives v = 2.252 m/s

now v=u+at

where v=2.252 , u=0,a=6.036

gives t = 0.3731s

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