Question

You are skiing and have a mass of 47 kg and are on a slope making an angle of 37 degrees above horizontal. The coefficient of kinetic friction between the skis and the snow is 0.18.
What is the amount of force pulling you down the slope?
What is your acceleration down the slope?

Answer 1

gravity down slope=mg*SinTheta
forcepulling-friction=mass* acceleration
mgSinTheta-mu*mgCosTheta=mass*acceleration
solve for acceleration

Answer 2

Is this right?
(45.0)*(9.80)*(sin35 degrees)= a number
a number - frictional force = answer

Answer 3

yes. but you said 37 deg, not 35

Answer 4

Oh sorry. To find the frictional force is it
(0.18)*(47.0)(9.80)cos37 degrees

and the magnitude of the Normal force is
(47.0)(9.80)cos37 degrees

Answer 5

yes

Answer 6

For the frictional force I got 61.4
For the magnitude of the Normal force I got -398.5
and for the amount of force pulling you down the slope I got -250.2
I don't think that's right though.

Answer 7

(45.0)*(9.80)*(sin35 degrees)= a number
45kg or 47kg?

Answer 8

Sorry I meant 47 kg.

Answer 9

pulling force=mg*sin37=45*9.8*1/2 about
pulling force= about 225 in my head. how did you get -250?
In my head, frictional force=about 65
Now let me simplify a lot of this. Notice mass divides out..

mgSin37-mu*mg*cos37=ma
g(sin37-muCos37)=a

in my head
9.8*about(1/2- 1/5*.796)=
9.8*about (.5-.14)=9.8*.36 about 3.5m/s^2\
check all that.

Answer 10

If you used 47kg and 37 degrees, you need to redo your calculations.
Check your calculator for the trig functions to see if it is set in degree mode.
In degree mode: cos(60)=0.50000000
In radian mode: cos(60)=-0.95241298...

Answer 11

Ohh I see the pulling force is 220.5
The frictional force is 66
For the normal force I actually got 367.85

I just don't understand how you solved for acceleration by saying 1/2-1/5*7.96