Question

A block of mass m = 3.5 kg is attached to a spring with spring constant k = 990 N/m. It is initially at rest on an inclined plane that is at an angle of θ = 22° with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is μ_k = 0.12. In the initial position, where the spring is compressed by a distance of d = 0.19 m, the mass is at its lowest position and the spring is compressed the maximum amount. Take the initial gravitational energy of the block as zero.

what is the blocks initial mechanical energy in joules? If the spring pushes the block up the incline, what distance, L in meters, will the block travel before coming to rest? the spring remains attached to both the block and the fixed wall throughout its motion

Answer 1

Ans:-

Given data:-Mass of block m= 3.5kg

Spring constant k= 990N/m

Initially at rest means vi = 0m/s

Θ = 22de,

Kinetic friction = μk = 0.12

Compressed distance of d = 0.19m

1 we want to find out initial mechanical energy

ME initial = PE + KE

                = mgh + ½ kx^2 + ½ mvi^2

H=0m

X= d = 0.19m

Vi = 0m/s

So

ME initial = 0+ ½ kd^2 + 0

                = ½ *990*0.19^2

                = 17.87j

Conservation of mechanical energy

MEi = MEf

MEi = KE + PE    

17.87 = 0 + mgLsin22 – μkmgLcos22

17.87 = (3.5*9.8sin22 – 0.12*3.5*9.8*cos22)*L

17.87 = (12.85 – 3.82)*L

L = 1.98m