# Homework Help: Why can't i do it my way

1. May 6, 2010

### xstetsonx

A 400 g particle moving along the x-axis experiences the force shown in Figure Ex11.15. The particle goes from vx = 2.5 m/s at x = 0 m to vx = 5.5 m/s at x = 2 m. What is Fmax?

my solution manual told me to do this with work and energy

but why can't i do
kinematic:
5.5^2=2.5^2+2A2=6

F=ma
(0.4KG)(6m/s^2)=.5Fmax(2)=2.4N
area under the curve--triangle---

but the correct answer is 4.8N.........ugh

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2. May 6, 2010

### xstetsonx

Wait the area underneath the curve is work right and my answer is the total force? ..... so i have to do W=(F)(delta X)???????

3. May 6, 2010

### willem2

If the curve had the force as a function of time then your approach would work.

$$m(v_2-v_1) = \int_{t_1}^{t_2} F(t) dt$$

this is simply F=ma integrated

but since it isn't you can't do this and you will have to use

$$W = \int_{x_1}^{x_2} F(x) \cdot dx$$