Draw a FBD, calculate tension in the string, determine mass

• jfnn
I also got .93 for the tension and 3.72 for the mass. Thank you for the reassurance!In summary, the conversation discusses a problem involving a box sliding down a ramp with friction and then sliding on a smooth surface. The law of conservation of energy is used to calculate the final speed of the box as it slides on the floor. There are some minor typos and discrepancies in the calculations, but the general logic and final values for tension and mass are confirmed by multiple individuals.
jfnn

Homework Statement

A box slides on a smooth (frictionless) horizontal surface, 1.50 m above the floor, at a speed of 2.00 m/s. The box then slides down a ramp that makes an angle of 36 deg with the horizontal and has a coefficient of kinetic friction equal to 0.430. After reaching the end of the ramp, the box continues to slide horizontally on a smooth ground. Use the law of conservation of energy to calculate the final speed (v) of the box as it slides on the floor.

Homework Equations

[/B]
KEi + PEi = KEf + PEf

and

KEi + PEi + Wnc = KEf + PEf

The Attempt at a Solution

[/B]
I broke it into three segments. I first did the flat floor above the ramp. Since it is all conservative forces acting, I used KEi + PEi = KEf + PEf. Furthermore, down the ramp i defined as negative and south is negative.

Therefore,

1/2m*vi^2 + mghi = 1/2m*v^2 = mgh

In the above equation, the height initial (hi) and the final height (h) are both the same. So the mgh will cancel leaving

1/2m*vi^2 = 1/2m*v^2

After simplifying,

vi^2=v^2 --> Since vi is -2.00 m/s (it is going down in the direction towards the ramp, which is to the left_

(2.00)^2 = v^2
4=v^2
v=-2.00 m/s --> This will now become the initial velocity for the segment involving the ramp.

Since the ramp has friction, we need to consider work by the non-conservative forces -->

KEi + PEi + Wnc = KEf + PEf

1/2m*vi^2 + mghi + Wnc = 1/2m*v^2 + mgh

Where: The initial velocity (vi) = -2.00 m/s, the initial height (hi) is 1.5m, the final height (h) is 0, and wnc= dfcos(theta).

1/2m*(-2)^2 + m(9.8)(1.5) + d*friction*cos(theta) = 1/2m*v^2 + m(9.8)(0)

Therefore, 16.7m - d*uk*N = 1/2m*v^2

The distance (d) is: sin(36) = 1.5 --> d = 2.5519m

16.7m - (2.5519m)(0.430)*N = 1/2m*v^2

16.7m - 1.0973N = 1/2m*v^2

I use a force diagram of the box on the ramp, and I see that N=w perpendicular --> N= mgcos(36)

33.4m - mgcos(36) * 2.1946 = v^2

33.4 - 17.3995 = v^2

16=v^2

v= +/- 4.00 m/s --> I use negative value, again because I defined down the ramp as negative direction for x

Finally, just to show that this is the final velocity of the box, I take the segment of the box just hitting the ground where only conservative forces act, to show that the final velocity as it is sliding on the floor is -4.00 m/. This question asks one for speed, which is scalar, so the answer is 4.00 m/s. Is this the correct logic for a problem like this? Could someone verify the answer?

Attachments

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The logic looks fine for the ramp problem. There were a couple of typos, like when you had: 33.4m - mgcos(36) * 2.1946 = v^2

It should be 33.4m - mgcos(36) * 2.1946 = mv^2.

Last edited:
jfnn
scottdave said:
The logic looks fine for the ramp problem. There were a couple of typos, like when you had: 33.4m - mgcos(36) * 2.1946 = v^2

It should be 33.4m - mgcos(36) * 2.1946 = mv^2.

Thank you, i got 4 but another friend of mine got 7? Do you know if my math is right?

For the picture I attached I got the answers to be T= 0.93 N

I got the mass of cylinder to be 3.72 kg..

Does that seem right? Furthermore, for the FBD for the block it should only be Tension upwards and weight downwards? And then the cylinder should have both tension and weight downwards? Does that make sense?

Thank you again!

jfnn said:
the FBD for the block it should only be Tension upwards and weight downwards? And then the cylinder should have both tension and weight downwards? Does that make sense?

Thank you again!
So the cylinder turns and has a moment of inertia. Do you know how to handle those type of problems?

jfnn
scottdave said:
So the cylinder turns and has a moment of inertia. Do you know how to handle those type of problems?

Yup. I got the mass from that moment of inertia using torque = I * the angular acceleration.

Do you know how the free body diagram should look for the cylinder? Is it tension and weight pointing down?

jfnn said:
Yup. I got the mass from that moment of inertia using torque = I * the angular acceleration.

Do you know how the free body diagram should look for the cylinder? Is it tension and weight pointing down?
OK. Yes tension is down, weight is down, and reaction force at the axle is up.

jfnn
scottdave said:
OK. Yes tension is down, weight is down, and reaction force at the axle is up.

Thank you so much. Did you happen to get the same tension and mass by any chance? Some reassurance would be awesome. LOL

jfnn said:
Thank you, i got 4 but another friend of mine got 7? Do you know if my math is right?
I also came up with 4 m/s.
jfnn said:
For the picture I attached I got the answers to be T= 0.93 N

I got the mass of cylinder to be 3.72 kg..
I arrived at the same values for the tension and the cylinder mass. I wasn't sure if you need to know the radius, but it turns out that the radii cancel, so two cylinders of vastly different size but the same mass (say one made of styrofoam and the other made of lead) would allow the falling mass to behave the same.

jfnn
scottdave said:
I also came up with 4 m/s.

I arrived at the same values for the tension and the cylinder mass. I wasn't sure if you need to know the radius, but it turns out that the radii cancel, so two cylinders of vastly different size but the same mass (say one made of styrofoam and the other made of lead) would allow the falling mass to behave the same.

Thank you so much! I thought the same thing as you, which is very interesting!

What is a free body diagram (FBD)?

A free body diagram is a visual representation of the forces acting on an object. It shows the object as a point and all the forces acting on it, including their directions and magnitudes.

How do you draw a free body diagram?

To draw a free body diagram, first identify the object and draw it as a point in the center of your diagram. Then, draw arrows representing all the external forces acting on the object, labeling each arrow with the type of force and its direction. Make sure to include all forces, such as gravity, tension, and normal force.

How do you calculate tension in a string?

To calculate tension in a string, you need to use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration. In the case of a string, tension is equal to the force applied to it, which can be calculated by multiplying the mass of the object by its acceleration.

What factors affect the tension in a string?

The tension in a string is affected by the mass of the object it is supporting, the acceleration of the object, and the angle at which the string is being pulled. Additionally, the type and strength of the material the string is made of can also affect its tension.

How do you determine the mass of an object using a free body diagram?

To determine the mass of an object using a free body diagram, you need to first identify all the forces acting on the object and their magnitudes. Then, use Newton's second law to set up an equation with the known forces and the unknown mass, and solve for the mass. Alternatively, if you know the tension in the string and the acceleration of the object, you can use the formula for tension mentioned earlier to calculate the mass.

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