Answer:
Explanation:
Given that,
Spring constant k=200N/m
Compression x = 15cm = 0.15m
Attached mass m =2kg
Coefficient of kinetic friction uk= 0.2
The energy in the spring is given as
U =½kx²
U = ½ × 200 × 0.15²
U = 2.25J
Force in the spring is given by Hooke's law
F = ke
F = 200×0.15
F = 30N
The weight of body which is equal to the normal is give as
W = mg
W = 2 × 9.81
W = 19.62N
W = N = 19.62 Newton's 2nd Law
From law of friction,
Fr = uk•N
Fr = 0.2 × 19.62
Fr = 3.924
Using newton second law again
Fnet = F - Fr
Fnet = 30 - 3.924
Fnet = 26.076
Work done by net force is given as
W = Fnet × d
W = 26.076d
Then, the work done by this net force is equal to the energy in the spring
W = U
26.076d = 2.25
d = 2.25/26.076
d = 0.0863m
Which is 8.63cm
So the box will slide 8.63cm before stopping
