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Question: A trunk of mass 21 kg is on the floor. the trunk has a very small initial speed. the acceleration of gravity i?

(Posted by: JEN JEN on 2010-03-06 20:20:25)

What constant horizontal force pushing the trunk is required to give it a velocity of 10 m/ s in 20 s if the coefficient of sliding friction between the trunk and the floor is 0.58? Answer in units of N.

Posted by: perryinjax on 2010-03-06, 20:21:20

Man get off YA and go do your own homework!

Posted by: Somu on 2010-03-06, 20:43:19

Vi = 0 m/ s vf = 10 m/ s t = 20 s a = (vf - vi)/ t = (10 - 0)/ 20 = 0.5 m/ s^2 Fnet = m*a = 21 * 0.5 = 10.5 N Or Fapplied - Ffriction = 10.5 N Fapplied = Ffriction + 10.5 N = umg + 10.5 N = 0.58 * 21 * 9.8 + 10.5 N = 130 N Ans: 130 N ____________________.

Posted by: jcherry_99 on 2010-03-06, 20:50:00

Ff = force of friction = mu*m*g Ff = ?? mu = 0.58 m = 21 kg g = 9.8 Ff = 0.58*21 * 9.8 Ff = 119.5 vi = 0 very small vf = 10 m/ s t = 20 sec. a = (vf - vi)/ t = 10/ 20 = 0.5 m/ s^2 m = 21 kg Fr = m*a F - Ff = Fr F - Ff = 21*0.5 = 10.5N F - 119.5 = 10.5 F = 119.5 + 10.5 = 130 N The force you have to apply is 130 N