A block of mass,m,is pushed up against a spring,compressing it a distance x,and is then released.The spring projects the block along a frictionless horizontal surface,giving the block a speed v.The same spring projects a second block of mass 4m,giving it a speed 3v.What distance was the spring compressed in the second case?
A)x
B)2x
C)3x
D)4x
E)6x

Question

Quizplus · Accepted Answer

Let's first use the conservation of energy to find the initial compression of the spring in the first case. The potential energy stored in the spring when it's compressed by x is:

U = 1/2 k x^2

where k is the spring constant. When the block is released, all the potential energy is converted to kinetic energy:

1/2 m v^2 = U = 1/2 k x^2

Solving for x, we get:

x = sqrt(m v^2 / k)

In the second case, the same spring is used to project a block of mass 4m with speed 3v. Again, using the conservation of energy, we have:

1/2 (4m) (3v)^2 = U = 1/2 k y^2

where y is the unknown compression of the spring in the second case. We can rewrite this as:

y^2 = 36 (m/k) v^2

Now we can use the ratio of x and y to find the answer:

y/x = sqrt(36 (m/k) v^2 / (m v^2 / k))
y/x = 6

Therefore, the distance the spring was compressed in the second case is 6 times the distance in the first case, which is 3x. Hence, the answer is C) 3x.

A Block of Mass,m,is Pushed Up Against a Spring,compressing It