# Forces and Motion Questions

#### EddieV

1. The problem statement, all variables and given/known data
The pulley device is used to hurl projectiles from a ramp (Coefficient of Kinetic friction [mu] = 0.26) as illustrated in the diagram. The 5.0 kg mass is accelerated from rest at the bottom of the 4.0 m long ramp by a falling 20.0 kg mass suspended over a friction less pulley. Just as the 5.0 kg mass reaches the top of the ramp, it detaches from the rope (neglect the mass of the rope) and becomes projected from the ramp.

The ramp angle is 30 degrees

a) Determine the acceleration of the 5.0 kg mass along the ramp.

b) Determine the tension in the rope during acceleration of the 5.0 kg mass along the ramp.

c) Determine the speed of the projection of the 5.0 kg mass from the top of the ramp.

d) Determine the horizontal range of the 5.0 kg mass from the base of the ramp.

2. Relevant equations
Let m1 = 5.0 kg
Let m2 = 20.0 kg

3. The attempt at a solution
For a) I know how to solve the acceleration of the system with
(m2g - 0.26*m1g*cos30 - m1g*sin30)/(5+20)
So to find the acceleration of just the 5.0 kg block do I use this formula except only divide by 5 instead of 5 + 20. Or is the acceleration of the system also the acceleration of this block?

For b) Im pretty sure you do the acceleration of the system * m1 + Force of Kinetic friction which is 0.26*m1g*cos30

For c) I was thinking of using the big 5, knowing initial velocity is 0, displacement is 4.0 m, and either acceleration of the system or the accelerated I found in a) [unless they're the same acceleration]

For d) I would solve it as a projectile motion problem, I haven't really given it that much thought though

I feel most of my issues boil down to not knowing if the acceleration of the system is the same as the acceleration of the each block independently, and if they are separate values then how do you solve for them.

Thank you and have a nice day!

#### RPinPA

So to find the acceleration of just the 5.0 kg block do I use this formula except only divide by 5 i
That is the total force acting on everything. The system moves together as one unit. If the 20 kg mass goes down 1 cm, the 5.0 kg mass goes up 1 cm. The length of the rope by assumption never changes, so they can't have different motions.

For b) Im pretty sure you do the acceleration of the system * m1 + Force of Kinetic friction which is 0.26*m1g*cos30
That does not represent all the forces acting on m1.

For c) I was thinking of using the big 5,
Meaning the standard kinematic equations for uniformly accelerated motion? That should work. I usually try energy considerations first as it's often easiest, but that leads to some of the same equations.

For d) I would solve it as a projectile motion problem, I haven't really given it that much thought though
Reasonable, since once it leaves the top of the ramp it's a projectile. But carefully read what distance was asked for.

#### haruspex

Homework Helper
Gold Member
2018 Award
If the 20 kg mass goes down 1 cm, the 5.0 kg mass goes up 1 cm.
To clarify, the 5.0 kg mass goes up the slope 1 cm.

#### EddieV

Thank you for the help I think I understand a) a lot better for b) though is the force that I'm missing the one that is parallel to the slope of the ramp and a component of the m1g force?

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