What is its speed after traveling 1.00 m on the rough surface?

[cc21392001]

physics homework help ASAP thx!

thx so much, here they are:

48. A block of mass 0.6 kg slides 6.0 m down a frictionless ramp inclined at 20 degrees to the horizontal. It then travels on a rough horizontal surface where coefficient of kinetic friction is 0.50.
a) what is its speed after traveling 1.00 m on the rough surface?
b) what distance does it travel on this horizontal furface before stopping?

68. An electron moves with a speed of 0.995c.
a) what is its kinetic energy?
b) if you use the classical expression to calculate its kinetic energy, what percentage error would result?

k = mc^2((1/(1-(v/c)^2))-1)

thx a lot

 

[drag]

Greetings !

First of all welcome to PF ! And, as a member, you should
know that there are certain guidelines in every forum
here and the main guideline of this forum is that you
have to show your attempt at a solution and not just
ask for one.

I'll give you some general hints though - in the first
question you'd better use energy calcs. Find the energy
at the bottom and think how you can compare it to the
energy lost due to friction on the surface.

In the second, well, just use the formula after you
understand what its parts mean.

Live long and prosper.

 

[cc21392001]

oh ok

well for the first one i found the frictional force that's retarding the energy of the object, so that's friction coefficient x normal force x distance, 0.5 x 0.6 x 9.8 x cos20 = 2.76, the distance is one.
then i subtract the total work which is 0.6 x 9.8 x 6sin20 = 12.07 with 2.76 and get 9.31 J. ok, now wut happens? i spent about 3 hours on this problem already, please help!

 

[arcnets]

Originally posted by cc21392001
well for the first one i found the frictional force that's retarding the energy of the object, so that's friction coefficient x normal force x distance

Correct, but that's happening on the horizontal part, not the ramp, right? Please redo.

then i subtract the total work

OK.

ok, now wut happens?

You should use the formula for kinetic energy.
$$T = \frac{1}{2}mv^2.$$

 

[krab]

Originally posted by cc21392001
oh ok

well for the first one i found the frictional force that's retarding the energy of the object, so that's friction coefficient x normal force x distance, 0.5 x 0.6 x 9.8 x cos20 = 2.76...

The ramp is frictionless...

..., the distance is one.
then i subtract the total work which is 0.6 x 9.8 x 6sin20 = 12.07 with 2.76 and get 9.31 J. ok, now wut happens? i spent about 3 hours on this problem already, please help!

You've spent 3 hours on it and still didn't realize the ramp is frictionless? Maybe your problem is English.

 

[cc21392001]

naw man the problem says "on the rough surface"