A constant horizontal pull acts on a sled on a horizontal frictionless ice pond.The sled starts from rest.When the pull acts over a distance x,the sled acquires a speed v and a kinetic energy K.If the same pull instead acts over twice this distance:
A) The sled's speed will be 2v and its kinetic energy will be 2K.
B) The sled's speed will be 2v and its kinetic energy will be K
C) The sled's speed will be v and its kinetic energy will be 2K.
D) The sled's speed will be v and its kinetic energy will be K
E) The sled's speed will be 4v and its kinetic energy will be 2K.

Question

A constant horizontal pull acts on a sled on a horizontal frictionless ice pond.The sled starts from rest.When the pull acts over a distance x,the sled acquires a speed v and a kinetic energy K.If the same pull instead acts over twice this distance:
A) The sled's speed will be 2v and its kinetic energy will be 2K. 
B) The sled's speed will be 2v and its kinetic energy will be K 
C) The sled's speed will be v and its kinetic energy will be 2K. 
D) The sled's speed will be v and its kinetic energy will be K 
E) The sled's speed will be 4v and its kinetic energy will be 2K.

Quizplus · Accepted Answer

The work done by the constant horizontal pull is equal to the change in kinetic energy of the sled. Therefore, for the same pull acting over twice the distance, the work done (and hence the change in kinetic energy) is doubled. This means that the final kinetic energy of the sled will be 2K. 
Using the work-energy theorem: 
Work done = Change in kinetic energy 
Fxd = (1/2)mv^2 
Where F is the constant horizontal pull, x is the distance over which the pull acts, m is the mass of the sled, and v is the speed acquired by the sled. 
For the same pull acting over twice the distance, the work done will be 2Fx, and the final kinetic energy will be 2K: 
2Fx = (1/2)mv^2 
2(2Fx) = (1/2)m(2v)^2 = 2K 
Solving for v, we get: 
2Fx = (1/2)mv^2 
4Fx = mv^2 
v = sqrt(4Fx/m) 
For the pull acting over twice the distance, the distance is 2x, so: 
v' = sqrt(4F(2x)/m) = sqrt(8Fx/m) = sqrt(2(4Fx/m)) = sqrt(2)v 
So the sled's speed will be v/sqrt(2), which is approximately 0.7v. Therefore, the answer is C.

A Constant Horizontal Pull Acts on a Sled on a Horizontal