Homework Help: Force and energy problems

1. Nov 18, 2008

nns91

1. The problem statement, all variables and given/known data

1.A 6kg block slides down a frictionless incline making an angle of 60 degree with the horizontal.(a) What is the speed of the block after it has slid 1.5 if it starts from rest. (b) What is its speed after 1.5m if it starts with an initial speed of 2 m/s ?

2. A single force of 5 N acts in the x direction on an 8kg object. (a) If the object starts from rest at x=0 at time t=0, find its velocity v as a function of time.(b) write an expression for the power input as a function of time.(c) What is the power input of the force at time t= 3s

2. Relevant equations

W=KE
F=ma

3. The attempt at a solution

1. I use W=Kinetic energy to find v. I found down that Wtotal= Wgravity + Wnormalforce=102N.

Then I set 102=1/2 mv^2. But I got marked wrong. Any opinion ??

2. I use F=ma to find the acceleration then use acceleration as the slope of V(t) but I am wrong again. How should I do this ?

2. Nov 18, 2008

nasu

For 1.
If by "W" you mean kinetic energy, then Wgravity and Wnormalforce are meaningless.
And the energy is not measured in N but in joules.
I suppose you mean to write conservation of energy or work-energy theorem.
Either one will work. The normal force does not do any work (it's normal, right? this means perpendicular)
So, for example the work-energy theorem:
KE_final = KE_initial +Work
Work=mgh (work done by gravity)

For 2.
It seems OK so far. Show your work so we can see where is the mistake.
I assume that by using the acceleration as the slope of V(t) you mean V(t)=a*t.

3. Nov 19, 2008

nns91

2. So here is my work:

F=ma so a=F/m=5/8=0.625 m/s^2

v(t)=0.625t

Is that right ?

4. Nov 19, 2008

nns91

1. That's what I meant to say. So did I do it right or wrong ?

I got W=KE so 102=(mv^2)/2 and solve for v

5. Nov 19, 2008

nasu

Yes, so far it seems OK to me.

6. Nov 19, 2008

nasu

What is 102? From where do you get it ?

7. Nov 19, 2008

nns91

It is the gravitational force's work

8. Nov 19, 2008

nasu

You mean m*g*h?
m=5kg, g=10m/s (aprrox); That will give h=about 2 m??
It slides 1.5 m along the plane. What is the vertical distance corresponding to this?
Use the angle of the incline. Cannot be MORE than 1.5 m.

9. Nov 19, 2008

nns91

no.

I got W=m*g*sin60*2.

2 is the distance it slides, not the vertical

10. Nov 19, 2008

nasu

In the problem it says 1.5 m. So why 2?

When you multiply by Sin(60) you get the vertical distance. But it should be 1.5*Sin60.
I think you mixed the distance with the speed (2m/s).