How Much Kinetic Energy Would the Ice Block Gain Without the Rope?

[Dan Feerst]

Homework Statement

a block of ice slides down a frictionless ramp at angle Θ = 52.0° while an ice worker pulls on the block (via a rope) with a force that has a magnitude of 52.0 N and is directed up the ramp. As the block slides through distance d = 0.8 m along the ramp, its kinetic energy increases by 82.0 J. How much greater would its kinetic energy have been if the rope had not been attached to the block?

Homework Equations

The Attempt at a Solution

Here is what I have come up with so far. Now I may be totally wrong on this but I think that since this senerio is frictionless, if we set up a coordinate plane with the x-axis = to Θ and F_N-(mg)i=0 => F_N=(mg)i
and W_net=F_r-(ma)jcos(0) we can treat this situation is if the fall was vertical.
Based on this, It seems like the problem should be as simple as adding the energy lost by the tension to the given 82.0J. which would be Fdcos(180)= -41.6
Here is where I'm getting screwed up because I think I'm setting something up backwards. I tried to reason it and plug in 123.6 for an answer but this is not correct either.
what am I doing wrong?

 

[anathema]

Thank you for your question. It seems like you have a good understanding of the problem so far. Let's break it down step by step.

First, let's consider the forces acting on the block. We have the force of gravity (mg), the normal force (FN), and the tension force from the rope (T). Since the ramp is frictionless, we can ignore the force of friction.

Next, let's set up a coordinate system where the x-axis is parallel to the ramp and the y-axis is perpendicular to the ramp. This will make our calculations easier. We can also use the fact that the block is sliding down the ramp at a constant speed, which means that the net force in the x-direction is 0.

Using Newton's second law, we can write the following equations:

∑Fy = T - mgcosΘ = 0 (since the block is not moving in the y-direction)
∑Fx = mgcosΘ - TsinΘ = ma (since the block is moving at a constant speed)

Solving for T in the first equation, we get T = mgcosΘ. Substituting this into the second equation, we get:

mgcosΘ - mgcosΘsinΘ = ma
a = gsinΘ

Now, let's consider the work done by the tension force. Since the block is moving at a constant speed, the work done by the tension force will equal the change in kinetic energy of the block. In other words:

W = ∆K = 82 J

Now, we can use the work-energy theorem to relate the work done by the tension force to the change in kinetic energy:

W = Fdcos180 = -Fd

Substituting this into our equation for work, we get:

-Fd = 82 J
d = -82 J/F = -0.16 m

Therefore, if the rope had not been attached to the block, the block would have traveled a distance of 0.16 m down the ramp before coming to a stop. This means that the kinetic energy of the block would have been:

K = 1/2mv^2 = 1/2m(vf^2 - vi^2) = 1/2m(0 - 2gdsinΘ) = mgsinΘd = 52.0 N * -

 

[tomcue]

I would like to clarify a few things before providing a response to this content. Firstly, the statement mentions a change in kinetic energy, but the given values and equations seem to be related to work and not kinetic energy. Secondly, the problem states that the block is sliding down a ramp, but then the attempt at a solution mentions a vertical fall. It is important to make sure that the variables and equations used are applicable to the given scenario.

Assuming that the problem is related to work and not kinetic energy, the first step would be to calculate the work done by the tension force on the block. This can be done by multiplying the force (52.0 N) by the distance (0.8 m) and the cosine of the angle between the force and the displacement (180°). This would result in a negative work of -41.6 J, as mentioned in the attempt at a solution.

However, it is important to note that this work is done by the tension force on the block, not on the block itself. The work done on the block would be the opposite of this, which would be a positive work of 41.6 J. This work would contribute to the increase in the block's kinetic energy, resulting in a total kinetic energy of 123.6 J (82.0 J + 41.6 J).

If we were to consider the scenario where the rope was not attached to the block, there would be no external force doing work on the block. Therefore, the work done on the block would be zero and its kinetic energy would remain at 82.0 J. So, the kinetic energy would have been 41.6 J greater if the rope had not been attached.

In conclusion, the correct approach to solving this problem would be to calculate the work done by the tension force on the block and then use this work to determine the change in the block's kinetic energy. It is important to use the correct variables and equations that are relevant to the given scenario.