How Does Changing Force Angle Affect Kinetic Energy Increase in a Sled?

[Parzival]

Help! I am a starter physicist and I need hlep on this question stated thus:

Homework Statement

A sled is being pulled across a horizontal patch of snow.
Friction is negligible. The pulling force points in the same direction
as the sled's displacement, which is along the +x axis. As a result, the kinetic energy of the sled increases by 38%. By what percentage would the sled's kinetic energy have increased if this force had pointed 62° above the +x axis?

Homework Equations

W = final KE - initial KE
W = (F cos theta)*displacement

The Attempt at a Solution

I tried using the work-energy theorem, but i couldn't solve for any of the variables. Then I turned to the second work formula, but still no dice.

 

[azizlwl]

Since Friction is negligible, no work done.

 

[ehild]

Homework Statement

A sled is being pulled across a horizontal patch of snow.
Friction is negligible. The pulling force points in the same direction
as the sled's displacement, which is along the +x axis. As a result, the kinetic energy of the sled increases by 38%. By what percentage would the sled's kinetic energy have increased if this force had pointed 62° above the +x axis?

Homework Equations

W = final KE - initial KE
W = (F cos theta)*displacement

The Attempt at a Solution

I tried using the work-energy theorem, but i couldn't solve for any of the variables. Then I turned to the second work formula, but still no dice.

You have all the necessary formulae. The work is equal to the change of KE. Write it out for both cases. The angles are different, but the displacements same x.
Call the intial kinetic energy KE_i. Write the change of KE in terms of KE_i. It is 0.38KE_i in the first case, and ηKE_i in the second case when theta=62°. You need to find the unknown factor η.
Divide the equations with each other: Both KE_i and x cancel. Solve for η .



ehild

 

[ehild]

Since Friction is negligible, no work done.

That is not right. The sled is accelerated. The change of kinetic energy needs work.

ehild

 

[Kugan]

The first time I read the problem I did a different way than was suggested.

If the initial force parallel to the ground causes an inc in the Kinetic energy by 38% then a force at an angle of 62 degrees, there is only 46.9% of the original force being applied horizontally.

How I got 46.9% is by F cos(angle)= force in the x direction
Fcos(62) = ""
F 0.469 = "" ""
This is essentially saying that only .469% of the force is in the x direction( the rest in the y-direction).
In the original case(where there is 38% inc) the angle is zero and cos 0= 1 so 100% of the force is in the x direction.
So if 100 of the force = 38% , then 46.9 percent of the force = ?simple ratio 100 = 38
----- ----
46.9 x
Therefore what is x this is 17% so there is only a 17% inc in Kinetic energy when the force is applied at 62 degrees to the ground.

Let me know if you want the algebraic method suggested by the other poster you will still arrive at the same answer, but its longer.

 

[ehild]

Therefore what is x this is 17% so there is only a 17% inc in Kinetic energy when the force is applied at 62 degrees to the ground.

Let me know if you want the algebraic method suggested by the other poster you will still arrive at the same answer, but its longer.

Both methods are essentially the same.

ehild

 

[Parzival]

I figured it out by myself, but thank you all.
The ratio of the initial work to the work w/ the 62° angle is

Fs/s(f cos 62°)
= 1/(cos 62°)

Thus, multiply 38% by cos 62° (0.469).
Significant digits yield 0.18, or 18%.