Answer:
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
