# Homework Help: Physics help on force and speed

1. Jul 31, 2013

### bestchemist

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

FIGURE 7-6

The force on a 3.00-kg object as a function of position is shown in Fig. 7-6. If an object is moving at 2.50 m/s when it is located at x = 2.00 m, what will its speed be when it reaches x = 8.00 m?

________
A)

2.90 m/s

B)

3.30 m/s

C)

3.70 m/s

D)

4.10 m/s

E)

4.50 m/s

2. Relevant equations

Have no idea

3. The attempt at a solution
the answer is 3.3 m/s but I don't know how they got that answer

Last edited by a moderator: May 9, 2015
2. Jul 31, 2013

### PhanthomJay

You might want to use work energy methods and show an attempt, please. Thank you.

3. Jul 31, 2013

### bestchemist

I tried but my answer came out to be something that's not even on the answer choice lol
Here is what I did

Δk = w
1/2 mv^2 = Fd
(1/2)(3)v^2 = 1*8
V= 2.3 m/s

4. Jul 31, 2013

### barryj

First find an expression for the acceleration vs distance. Then integrate the acceleration to get the velocity vs distance. Use the fact that at x = 2 the velocity = 2.5 to find the constant of integration. Now you can find the velocity at x = 3. Now, knowing the velocity, mass and force at x = 3, you can find the velocity at x = 8.

5. Jul 31, 2013

### Staff: Mentor

You had the right idea, but you made a couple of mistakes. When you wrote the change in kinetic energy on the left hand side of the equation, you left out the term for the initial kinetic energy at x = 2. On the right hand side of the equation, you obtained the wrong value for the integral of the force with respect to x, from x = 2 to x = 8. The correct value should be 6.5.

Chet

6. Jul 31, 2013

### bestchemist

Got it!! Thank y'all :)

7. Oct 17, 2016

### NJIT Physics Tutor

• Member advised not to provide complete solutions to homework questions.
Change in K = W
1/2*m*(vf^2-v0^2) = F*d
.5*3.00kg*(vf^2-(2.5m/s)^2) = 1N * (8-2m)
1.5 (vf^2-(2.5m/s)^2) = 6N*m
vf^2 = 6/1.5 + (2.5 m/s)^2
vf^2 = 10.25
vf = 3.20 m/s