Block moving up an incline with friction

  • #1
249
1

Homework Statement


A block of mass m is moving along a horizontal frictionless surface with a speed of 5.7 m/s. The block encounters a slope which it travels up. If the slope is 11 degrees and the coefficient of kinetic friction between the block and the incline is 0.26, how far does the block travel up the incline?


Homework Equations


fk = uk*Fn

K1 + U1+ Wother = K2 + U2


The Attempt at a Solution



Using a free body diagram I found the normal force, Fn, to be

Fn - mgcos(11) =0
Fn = 2.50119

Then using K1 + U1 + Wother = K2 + U2

1/2*m*v^2 - 2.50119m = 9.8*m*h

I solved for h to find the height that the block travels up the ramp
h = 1.40243 m

Then using SOHCAHTOA I solved for the length of the ramp that the block traveled:
1.40243/sin(11) = 7.34992m

I know the answer should be 3.7 meters but I don't know where I'm going wrong.
 

Answers and Replies

  • #2
PeterO
Homework Helper
2,431
54

Homework Statement


A block of mass m is moving along a horizontal frictionless surface with a speed of 5.7 m/s. The block encounters a slope which it travels up. If the slope is 11 degrees and the coefficient of kinetic friction between the block and the incline is 0.26, how far does the block travel up the incline?


Homework Equations


fk = uk*Fn

K1 + U1+ Wother = K2 + U2


The Attempt at a Solution



Using a free body diagram I found the normal force, Fn, to be

Fn - mgcos(11) =0
Fn = 2.50119

Then using K1 + U1 + Wother = K2 + U2

1/2*m*v^2 - 2.50119m = 9.8*m*h

I solved for h to find the height that the block travels up the ramp
h = 1.40243 m

Then using SOHCAHTOA I solved for the length of the ramp that the block traveled:
1.40243/sin(11) = 7.34992m

I know the answer should be 3.7 meters but I don't know where I'm going wrong.

I think your "value" of Fn should have an m in it [you had it in later so probably just a typo] HOWEVER, 11 degrees is quite a slight slope, so I think the normal force should be much closer to that from a horizontal surface. not the full 9.8m, but probably not below 8m. Was your calculator set to degrees when you found cos(11)?

I would probably calculate how much energy is lost to friction and converted to Potential energy [mgh] for each metre travelled up the slope, then by comparison to the initial Kinetic energy. compute how many metres along the slope it would go.
 
  • #3
249
1
I've made a mistake with my labeling. Fn = 2.50119m is actually Fn*uk = fk. So I actually mean that fk = 2.50119.

I Fn = 9.61995m

If I labelled the distance that the block goes up the slope D, do I need to multiple my value of fk by D?
 
  • #4
cepheid
Staff Emeritus
Science Advisor
Gold Member
5,192
38
I've made a mistake with my labeling. Fn = 2.50119m is actually Fn*uk = fk. So I actually mean that fk = 2.50119.

I Fn = 9.61995m

If I labelled the distance that the block goes up the slope D, do I need to multiple my value of fk by D?

Yeah. The work done by friction is equal to the frictional force multiplied by the distance over which it acts. In this case, that's the distance along the incline that the object travels.
 
  • #5
249
1
I've got it now:

1/2*m*(5.7)^2 - 0.26*m*9.8*cos(11)*d= 9.8*m*dsin(11)
16.245 = 4.37112d

d=16.245/4.37112
d= 3.7 m

where d is the distance the block travels up the ramp.

Thanks for your help.
 

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