The Student Room Group

M1 November 2002 question 5.b)

I'm sure I'm being stupid, but I don't know where I'm going wrong on this question, could somebody help?

A suitcase of mass 10kg slides down a ramp which is inclined at an angle of 20 degrees to the horizontal. The suitcase is modelled as a particle and the ramp is a rough plane. The top of the plane is A. The bottom of the plane is C and AC is a line of greatest slope. The point B is on AC with AB=5m. The suitcase leaves A with a speed of 10 ms-1 and passes B with a speed 8 ms-1.

Find b) the coefficient of friction

So this is what I did:

to find deceleration of suitcase,
u=10 ms-1
a=8ms-1
s=5m
a=?
v^2=u^2 + 2as
64=100 + 10a
a= - 3.6 ms-2

Resolving perpendicular to plane,
R-10gsin20 = 0
R=10g sin20

Resolving parallel to plane,
-3.6 x 10 = 10gcos20 - F
F = 10 g cos 20 + 36

Plugging this into the formula Fmax = coefficient of friction x R
gives 3.82.

This answer, according to the mark scheme is 0.755. Please help!
redpanda
Resolving perpendicular to plane,
R-10gsin20 = 0
R=10g sin20

Resolving parallel to plane,
-3.6 x 10 = 10gcos20 - F
F = 10 g cos 20 + 36

Plugging this into the formula Fmax = coefficient of friction x R
gives 3.82.

This answer, according to the mark scheme is 0.755. Please help!
i've not completed the question yet, but i think R= 10gcos 20
Reply 2
redpanda
I'm sure I'm being stupid, but I don't know where I'm going wrong on this question, could somebody help?

A suitcase of mass 10kg slides down a ramp which is inclined at an angle of 20 degrees to the horizontal. The suitcase is modelled as a particle and the ramp is a rough plane. The top of the plane is A. The bottom of the plane is C and AC is a line of greatest slope. The point B is on AC with AB=5m. The suitcase leaves A with a speed of 10 ms-1 and passes B with a speed 8 ms-1.

Find b) the coefficient of friction

R= perpendicular force so F=uR
R=10gCos20

u=10 v=8 s=5 a=
64=100+10a
-36=10a
-3.6=a

f=ma
f=10x3.6
f=36

so 36/10gcos20=0.39 ???

have i done it wrong
Reply 3
darth_vader05
i've not completed the question yet, but i think R= 10gcos 20


Yeah thats correct.

R=10g cos 20
F=uR=u10g cos 20

Now resolving using F=ma
10g sin 20 - u10gcos20=10(-3.6) (i assumed a=-3.6 is correct from your first part)

10g(sin20-ucos20)=10(-3.6)
g(sin20-ucos20)=-3.6
ugcos20=gsin20+3.6
u=(gsin20+3.6) /gcos20
=0.755