# Newton's 2nd law question -- pushing a block up an inclined ramp...

1. Jul 2, 2015

### J-dizzal

If a constant force is applied to an object and the force is pushing the object up a inclined ramp with constant velocity what is the acceleration if mass is unknown?
F=ma
F=m (1m/s^2)​
If i apply this equation then acceleration is 1m/s^2. Am i applying this formula correctly? thanks

2. Jul 2, 2015

### Staff: Mentor

If the velocity is constant then what does that tell you about the acceleration?

3. Jul 2, 2015

### J-dizzal

i want to say if v is constant than acceleration is zero. but when i plug that into F=ma i got F=m0 which is zero force, but that doesnt work. and this is why im confused

4. Jul 2, 2015

### J-dizzal

because of the zero product property of multiplication

5. Jul 2, 2015

### Staff: Mentor

You're on the right track here - the acceleration is zero and therefore the net force is zero.

However, that zero net force is the sum of all the forces that are at play here, including gravity, friction, and the force being used to push the object up the ramp.

Last edited: Jul 2, 2015
6. Jul 2, 2015

### Staff: Mentor

If you think of Newton's 2nd law as $\Sigma F = ma$ (instead of just F = ma), that will remind you that it is the net force that equals ma.

7. Jul 2, 2015

### Noctisdark

Newton's second law is a way of saying, if you apply a force on a mass it will accelerate, F = ma describes the way it accelerate, you can see that the force F and acceleration a have the same direction, a = F/m mean that if if you apply a force on a very massive object it wil accelerate less than it would if you apply the same force on light object, in your case the object is moving at a constant velocity, so It isn't acceleration so the net force is zero as F = m*0 suggest, it's weight is pushing downward and the constant force you apply will cancel it but pushing upwards, good luck :p

8. Jul 2, 2015

### J-dizzal

yes i forgot about the sigma there. so in the example of an object on an inclined plane there would be a contact force applied to it, but when applying F=ma i must also add in the force that is parallel to the plane from its weight and frictional forces if they apply. F=ma would be ma=Fapplied-Fgravity-Ffriction, then can solve further for acceleration.
thanks!