# M1 questionWatch

#1
I have a question on F=ma:

A particle of mass 4kg is being pulled up a rough plane inclined at 25 to the horizontal by a force of magnitude 30N aciting along a line of greatest slope of the plane. Given that the particle is accelerating at 2m/s^2 find the coefficient of friction between the particle and the plane.

(I'm a bit unsure what to do with the 30N force?)
0
12 years ago
#2
(Original post by turtle2)
I have a question on F=ma:

A particle of mass 4kg is being pulled up a rough plane inclined at 25 to the horizontal by a force of magnitude 30N aciting along a line of greatest slope of the plane. Given that the particle is accelerating at 2m/s^2 find the coefficient of friction between the particle and the plane.

(I'm a bit unsure what to do with the 30N force?)
Resolve forces parallel to the plane

ma = 30 - friction - component of weight acting down the plane.

Then rearrange to find friction.

Then use f = mu*R to find mu (the coefficient of friction)

R = the component of weight perpendicular to the plane
0
12 years ago
#3
(Original post by turtle2)
I have a question on F=ma:

A particle of mass 4kg is being pulled up a rough plane inclined at 25 to the horizontal by a force of magnitude 30N aciting along a line of greatest slope of the plane. Given that the particle is accelerating at 2m/s^2 find the coefficient of friction between the particle and the plane.

(I'm a bit unsure what to do with the 30N force?)
30 is the resultant and it moves with acceleration 2 so we know that

30- resistance=ma

resolve forces
R=4x9.8cos25=35.5
F=4x9.8sin25=16.56
F=uR
16.56=35.5u
now bring in the 30
so 30-16.56-u35.5=8
35.5u=5.44
u=0.153

thanks latent corpse
0
12 years ago
#4
no mgsintheta is down slope and R=mgcostheta
0
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