Question

3. As shown in the figure below, a block weighing 14.0 N, which slides without friction on a 45.0° incline, is connected to the top of the incline by a massless spring of unstretched length 0.450 m and spring constant 120 N/m. (a) How far from the top of incline does the block stop i.e. reach equilibrium ? (b) If the block is pulled slightly down the incline and released, what is the time to reach the new equilibrium position (10 pts) | աաա 45.0

Answer 1

a)Forces on the mass along the incline include:
1. Component of weight along the incline=Wsin$\theta$ where W is weight of block,$\theta$ is angle of the incline with horizontal.

2.Note that since weight is acting down the incline, spring force must act up the incline for equilibrium.For that to happen, spring must be stretched. So,spring force=kx up the incline, where k is spring constant and x is amount by which spring is stretched from its natural length.

Since, block is in equilibrium, forces along the incline must be balanced to satisfy newton's second law.

So,kx=Wsin$\theta$=>x=Wsin$\theta$/k

Here,W=14 N, $\theta$ =45 degrees, k=120 N/m

So,magnitude of elongation x=14*sin45/120=0.0825 m

So,distance from top of the incline to equilibrium position=(natural length of spring)+(magnitude of elongation)=0.45+0.0825 m=0.5325 m = 53.25 cm

b)Note that for spring mass system,period of oscillation is given by:T=2$\pi$(m/k)^1/2,where T is period of oscillation of the simple harmonic motion, m is mass of block, k is spring constant.

Also, to go from extreme position to mean position , time taken is one-fourth of period of oscillation=1/4T

So,required time=1/4T=1/4[ 2$\pi$(m/k)^1/2 ]=$\pi$(m/k)^1/2/2.

Also,weight=mass*(gravitational acceleration).So, for the block,14=mass*(9.8), as gravitational acceleration=9.8 m/s²

=>mass=14/9.8=1.4286 kg

Also,spring constant k=120 N/m

So,required time=$\pi$(m/k)^1/2/2 =$\pi$(1.4286/120)^1/2/2 = 0.1714 seconds=0.17 seconds approximately