A box of mass m is pressed against (but is not attached to) an ideal spring of force constant k and negligible mass, compressing the spring a distance x. After it is released, the box slides up a frictionless incline as shown in the figure and eventually stops. If we repeat this experiment but instead use a spring having force constant 2k
A) the box will go up the incline twice as high as before.
B) just as it moves free of the spring, the kinetic energy of the box will be twice as great as before.
C) just as it moves free of the spring, the speed of the box will be times as great as before.
D) All of the above choices are correct.
E) None of the above choices is correct.

Question

A box of mass m is pressed against (but is not attached to) an ideal spring of force constant k and negligible mass, compressing the spring a distance x. After it is released, the box slides up a frictionless incline as shown in the figure and eventually stops. If we repeat this experiment but instead use a spring having force constant 2k 
A) the box will go up the incline twice as high as before. 
B) just as it moves free of the spring, the kinetic energy of the box will be twice as great as before. 
C) just as it moves free of the spring, the speed of the box will be times as great as before. 
D) All of the above choices are correct. 
E) None of the above choices is correct.

Quizplus · Accepted Answer

Let the initial compression of the spring be x, and let the angle of the incline be θ.

From conservation of energy, the gravitational potential energy lost as the box moves up the incline must be equal to the initial potential energy stored in the compressed spring:

mgx = (1/2)kx^2

Solving for x, we get:

x = 2mg/k

So, when we use a spring with force constant 2k, the initial compression will be:

x' = 2mg/(2k) = mg/k

Using conservation of energy again, we can find the maximum height h that the box will reach:

(1/2)mv^2 = mgh

Solving for h, we get:

h = v^2/(2g)

For the first experiment:

From conservation of energy, we know that the change in gravitational potential energy of the box is equal to the initial potential energy stored in the compressed spring:

mg(h+x) = (1/2)kx^2

Solving for h, we get:

h = (k/m)(x^2/(2g)) - x

Substituting x = 2mg/k, we get:

h = (2g/m)

For the second experiment:

The initial compression of the spring is mg/k, so the initial potential energy stored in the spring is:

(1/2)(mg/k)^2 = (1/2)(m/k)(g^2)

From conservation of energy, we know that the change in gravitational potential energy of the box is equal to the initial potential energy stored in the compressed spring:

mg(h+x') = (1/2)(2k)(mg/k)^2

Solving for h, we get:

h = 2g/m

So, in the second experiment, the box will go up the incline twice as high as before (choice A is correct).

To compare the kinetic energy and speed of the box in the two experiments, we can use conservation of energy again:

(1/2)mv^2 = mgh

For the first experiment:

v^2/2 = 2g(2g/m)

v^2 = 8g^2/m

For the second experiment:

v^2/2 = 2g(2g/m)(1/2)^2

v^2 = 2g^2/m

So, the kinetic energy of the box just as it moves free of the spring in the second experiment is half of that in the first experiment (choice B is correct).

Similarly, the speed of the box just as it moves free of the spring in the second experiment is 11eab6cc_1e15_302f_a957_f3a4ccf305ed_TB4469_11 times as great as in the first experiment (choice C is correct).

A Box of Mass M Is Pressed Against (But Is