Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
The coefficients of static and kinetic friction between a 476 N crate and the ware house floor are 0.615 abd 0.420, repectively...Asked by Sarah
The coefficients of static and kinetic friction between a 476 N crate and the ware house floor are 0.615 abd 0.420, repectively. A worker gradually increases his horizontal push against this crate utnil it just begins to move and from them on maintains that same maximum push. What is the acceleration of the crate after it has begun to move?
Netforce=ma
mg(mus-muk)=ma
solve for a
so I got 476 (9.8) (0.615-0.420) = ma
but what is the other m ???
Netforce=ma
mg(mus-muk)=ma
solve for a
so I got 476 (9.8) (0.615-0.420) = ma
but what is the other m ???
Answers
Answered by
drwls
There is only one mass, M, and it equals the weight divided by g, which is 48.6 kg.
The equation you wrote is the correct one after motion starts.
The force needed to start it moving is
Mg *(mu,stat) = 292.7 N, After it starts moving, the friction force becomes
Mg*(mu,kin) = 199.9 N
The net accelerating force is then
292.7 - 199.9 = 92.8 N. Divide that by M = 48.6 kg for the acceleration.
The equation you wrote is the correct one after motion starts.
The force needed to start it moving is
Mg *(mu,stat) = 292.7 N, After it starts moving, the friction force becomes
Mg*(mu,kin) = 199.9 N
The net accelerating force is then
292.7 - 199.9 = 92.8 N. Divide that by M = 48.6 kg for the acceleration.
Answered by
Sarah
so i got 1.91 is the right?
Answered by
shiv
absolutely right!
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.