Projectile Motion- SPH4U Grade 12 Physics

[AkadouYoroi]

Homework Statement

A pulley device is used to hurl projectiles from a ramp (uk=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 frictionless 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.
*there should be an attached image at the bottom*

a) Determine the acceleration of the 5.0-kg mass along the ramp. (Answer: 6.4 m/s2 up the ramp)

b) Determine the tension in the rope during the acceleration of the 5.0-kg mass along the ramp. (Answer: 68 N along the ramp)

c) Determine the speed of projection of the 5.0-kg mass from the top of the ramp. (Answer: 7.2 m/s)

d) Determine the horizontal range of the 5.0-kg mass from the base of the ramp. (Answer: 9.5m)

The Attempt at a Solution

a) I know that to find acceleration, i need to find the total net force of the system and divide by the two masses. I found the Fgx for the 5.0 kg object using (5.0)(9.0)cos30, the Fg for 20.0 kg object (20.0)(9.8), and the Ff of the system, using uFn. I don't get the quite answer. I am not sure if I missed any calculations, or my components are not correct.

b) I believe this might have been a component I've missed in part A, but I do not really know how to start off the tension calculation. I believe it involves using the frictional force to calculate the tension, but I am not sure.

c) I tried solving using the Vfy=V1y+ayt equation, and finding the value of t using y=1/2ay(t)2. (Sorry about not doing subscripts). I end up with an answer that's about 3.0s. But I think its because I might be confusing myself, because I've never done a projectile motion that involved mass in it.

d) Determine the horizontal range of the 5.0-kg mass from the base of the ramp.
Im not sure where to start, since it involves masses. I am not too sure if the object accelerated from rest in the question indicates something I should be doing.

Thank you so much.
The answers are posted with the questions, but I had a tough time getting to the answer, or not knowing how to start it.

 

[Doc Al]

a) I know that to find acceleration, i need to find the total net force of the system and divide by the two masses. I found the Fgx for the 5.0 kg object using (5.0)(9.0)cos30, the Fg for 20.0 kg object (20.0)(9.8), and the Ff of the system, using uFn. I don't get the quite answer. I am not sure if I missed any calculations, or my components are not correct.

The best way to attack this problem is by analyzing each mass separately. What forces act on the 5 kg mass? On the 20 kg mass? Apply Newton's 2nd law to each, which will give you two equations. That will allow you to solve for the two unknowns: tension and acceleration.

 

[ogb p]

Hello,

Thank you for providing the problem and your attempt at solving it. Projectile motion is a fundamental concept in physics and it is important to have a good understanding of it.

a) To find the acceleration of the 5.0-kg mass along the ramp, you need to consider the forces acting on the mass. The only forces acting on it are its weight, which is directed down the ramp, and the frictional force, which is directed up the ramp. Using Newton's Second Law, we can write:

ΣF = ma

Where ΣF is the sum of all the forces acting on the mass, m is the mass of the object and a is the acceleration. In this case, we can write:

ΣF = Fg - Ff

Where Fg is the weight of the mass and Ff is the frictional force. The weight can be calculated using Fg = mg, where g is the acceleration due to gravity (9.8 m/s^2). The frictional force can be calculated using Ff = μFn, where μ is the coefficient of friction (0.26 in this case) and Fn is the normal force, which is equal to the weight of the mass in this case. Therefore, we can write:

ΣF = mg - μmg = (1 - μ)mg

Substituting the values, we get:

ΣF = (1 - 0.26)(5.0)(9.8) = 34.6 N

Now, we can use Newton's Second Law to find the acceleration:

a = ΣF/m = 34.6/5.0 = 6.92 m/s^2

Note that this is the acceleration along the ramp, which is directed up the ramp.

b) To find the tension in the rope, we need to consider the forces acting on the 20.0-kg mass. The forces acting on it are its weight, which is directed down, and the tension in the rope, which is directed up. Using Newton's Second Law again, we can write:

ΣF = ma

Where ΣF is the sum of all the forces acting on the mass, m is the mass of the object and a is the acceleration. In this case, we can write:

ΣF = Fg - T

Where Fg is the weight of the mass and T is the tension in