# Force diagram for dynamics problem

1. Feb 5, 2006

### xc630

Hello i am having toruble with this problem.

A 24 kg block is pushed along a horizontal surface with an acceleration of 2.0 m/s^2. The coeficient of friction is 0.25 and teh block starts at 1.0m/s. Find: the force being applied horizontally, the velocity after 5.0 s, and the distance moved in 5.0 s.

I drew a force diagram by the way too.

Since the velocity is not constant
I have these premises: x axis is accelerated and the y- axis is at rest so it is at equilibrium so sigmaF=0

Thus I have

sigmaF= ma Sigmaup= sigma down
Sigma Fx = Fa(force applied) - f (friction) N(normal force)= Fw
Sigma Fx= Fa- uN N= Fw(f=uN)

What should I do next to find Fa the applied force? Thanks for any help

I don't know what to

2. Feb 5, 2006

### jamesrc

Set your equation for the sum of forces in the x-direction equal to the mass times the acceleration in the x-direction:

$$\Sigma_x F = ma = F_a - f$$
$$ma = F_a - \mu N$$
$$ma = F_a - \mu m g$$

I hope that helps.

3. Feb 5, 2006

### Hootenanny

Staff Emeritus
Well you know that the total horizontal force must be the sum of the acceleration force and the resistive force. Therefore,
$$F_x = ma + F_{friction}$$
$$F_x = ma + \mu R$$
$$F_x = ma + \mu mg$$ giving $$F_x = m(a + \mu g)$$

4. Feb 5, 2006

### xc630

Ok thanky you I got 530 N for the Fa, 11m/s for the velcoty and 130m for the distance. I have a question with this next problem though.

A 1500N block sits at rest on a ramp inclined at 40.0 degrees. The coeffeicients of starting and kinetic friction are 0.100 and there is a force of 250N applied up the ramp. I have to find the acceleration and net force. But how can I find the net force if I can't use sigmaFx= ma?

5. Feb 6, 2006

### Hootenanny

Staff Emeritus
You have to resolve the forces to their components parallel and perpendicular to the inclined plane. The apply $F = ma$ to find the acceleration where $F$ is the resultant force. You should still use the summation of forces, but be aware of the signs in front of the forces as some will be acting in opposite directions to the others.