A horizontal spring with spring constant 100 N/m is compressed 20cm and used to launch a 2.5kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the surface is 0.15.A. What is the speed of the box immediately after it loses contact with the spring?B. Find the change in kinetic energy of the box when it comes to a stop. C. Draw the free body diagram of the block when it is on the rough surface. D. Use work and energy to find how far the box slides across the rough surface before stopping.

Explanation:

Given that

Force constant k =100N/m

compression e=20cm =0.2m

Mass of box, =2.5kg.

a. Initial maximum compression K.E

Em₀ = K.E = ½ k x²

Final, free box without compressing the spring

 Emf =  K.E = ½ m v²

They are equal due to conservation of energy

    ½ k x² = ½ m v²

    v = √ (k / m) x

    v = √ (100 / 2.5) 0.20

    v = 1.265 m / s

b. Change in kinetic energy is given as,

∆K.E= Emf-Emo

Final point. Stopped box

Em{f} = 0

Starting point, starting the rough surface

Em₀ = K = ½ m v²

Therefore,

∆K.E=0-½×2.5×1.265²

∆K.E, =-2.J

∆K.E= -2J

c. Check attachment for free body diagrams

d. With Newton's second law we find the force of friction

fr = μ N

N-W = 0

N = W = mg

fr = μ mg

Work done by frictional force

Wfr} = -fr×d

W{fr} = - fr d

The workdone by friction is equal to the change in K.E

W(fr)=∆K.E

-μmgd=-2

d=2/μmg

d= 2/(2.5×9.8×0.15)

d=0.544m

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

false.

explanation:

this statement is false. after falling for 5 seconds, its speed will not be 100 m/s.