A2.5-kg box, sliding on a rough horizontal surface, has a speed of 1.2 m/s when it makes contact with a spring (see the figure. the block comes to a momentary halt when the compression of the spring is 5.0 cm. the work done by the friction, from the instant the block makes contact with the spring until is comes to a momentary halt, is -0.50 j.(a what is the spring constant of the spring?

The spring constant of the spring is 1040 N/m

Hooke's Law states that the length of a spring is directly proportional to the force acting on the spring.

F = Force ( N )

k = Spring Constant ( N/m )

Δx = Extension ( m )

The formula for finding Young's Modulus is as follows:

E = Young's Modulus ( N/m² )

F = Force ( N )

A = Cross-Sectional Area ( m² )

Δx = Extension ( m )

x = Initial Length ( m )

Let us now tackle the problem !

Given:

mass of the object = m = 2.5 kg

initial speed of the object = u = 1.2 m/s

final speed of the object = v = 0 m/s

compression of the spring = x = 5.0 cm = 0.05 m

work done by friction = W_f = -0.50 J

Unknown:

spring constant of the spring = k = ?

Solution:

We will use formula as follows:

Grade: College

Subject: Physics

Chapter: Elasticity

Keywords: Elasticity , Diameter , Concrete , Column , Load , Compressed , Stretched , Modulus , Young

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s