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

Further explanation

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:

Learn moreYoung's modulus : Young's modulus for aluminum : Young's modulus of wire :

Answer details

Grade: College

Subject: Physics

Chapter: Elasticity

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

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.

Then the answer would be d then