Physics Question ?

Question

A box slides down a ramp inclined at 35 degrees to the horizontal with an acceleration of 2.25 m/s2. The acceleration of gravity is 9.8 m/s2. Determine the coefficient of kinetic friction between the box and the ramp.

Answer 1

Good job Steve!

may I add some thoughts as to where it all came from

F=ma= Ft-uFn
where
F-the force of the object
Ft- Tangential to the surface force due to gravity
Fn- Normal to the surface force due to gravity
u - coefficient of kinetic friction
m- mass of the object
a - acceleration of the object
g- as you have said the gravitational acceleration

Ft= m g sin(35)
Fn=m g cos(35)

Then solving for u we have
u=(g sin(35) -a)/(gcos(35)

or as Steve has eloquently

u=(g sin(35) -a)/(gcos(35)=.42

Answer 2

We need to know the force of kinetic friction, f. Use the formula f=uR, where f is in N, u the coefficient we are looking for, and R is the force normal to the ramp. If the weight of the box is Mg then R =Mgcos35. It's the weight component normal to the ramp.

Thus f=uMgcos35.

We also have the weight component parallel to the ramp which is equal to Mgsin35. Obviously this is larger than f because it causes the box to accelerate at 2.25m/s^2.

Now use the basic formula: F=Ma where F is net force, M the mass of the box in kg, and a the given acceleration of 2.25m/s^2:

Substitute known values:

Mgsin35-uMgcos35=M*2.25 (M cancels out.)

9.8*0.574-u*9.8*0.819=2.25
5.62-8.03u=2.25
8.03u=5.62-2.25
8.03u=3.37
u=3.37/8.03
=0.42

Answer 3

here's a really simple equation to use when you need to find coefficient of kinetic, when only given Accleration and Degrees -

sin(Angle) - Cos(angle)(u) = A / g

so

sin(35) - cos(35)U = 2.25 / 9.81

0.573 - 0.819U = 0.229

-0.819U = 0.229 - 0.573

-.819U = -0.344

U = -0.344 / -0.819

U = 0.42

Answer 4

Hello - Nicole M - These guys sure answer lots of questions. This sort of thing used 2 b fun in Hi scol & college. Keep it up.

Answer 5

mu = tan 35 - a/(g cos35) = ~ .42