Question

A brick of mass m = 0.21 kg is set against a spring with a spring constant of k1 = 690 N/m which has been compressed by a distance of 0.1 m. Some distance in front of it, along a frictionless surface, is another spring with a spring constant of k2 = 489 N/m.
(A) How far d2, in meters, will the second spring compress when the block runs into it?
(B) How fast v, in meters per second, will the block be moving when it strikes the second spring?
(C) Now assume friction is present on the surface in between the ends of the springs at their equilibrium lengths, and the coefficient of kinetic friction is ?k = 0.5. If the distance between the springs is x = 1 m, how far d2, in meters, will the second spring now compress?

Answer 1

Answer:

Part a)

$x = 0.12 m$

Part b)

$v = 5.73 m/s$

Part c)

$x = 0.11 m$

Explanation:

Part a)

As per energy conservation we can say

Energy stored in spring 1 = energy stored in spring 2

$\frac{1}{2}k_1x_1^2 = \frac{1}{2}k_2x_2^2$

now we have

$k_1 = 690 N/m$

$x_1 = 0.1 m$

$k_2 = 489 N/m$

$x_2 = ?$

now from above equation

$(690)(0.1^2) = 489(x^2)$

$x = 0.12 m$

Part b)

To find the speed of the block we can say that its kinetic energy is gained due to spring energy

so we have

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

now we have

$\frac{1}{2}(690)(0.1^2) = \frac{1}{2}(0.21)v^2$

$v = 5.73 m/s$

Part c)

When friction is present between two springs then we have energy loss due to friction force

so we have

$\frac{1}{2}k_1x_1^2 - \mu mgd = \frac{1}{2}k_2x_2^2$

now we have

$\frac{1}{2}(690)(0.1)^2 - (0.2)(0.21)(9.8)(1) = \frac{1}{2}(489) x^2$

$3.45 - 0.4116 = 244.5 x^2$

$x = 0.11 m$