Question

A spring (70 N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41 ? with respect to the horizontal. The spring is then released.  (Figure 1)

Part A If the mass is not attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest? Express your answer using two significant figures. d =   m   SubmitRequest Answer Part B If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest? Express your answer using two significant figures. d =   m   SubmitRequest Answer Part C Now the incline has a coefficient of kinetic friction ?k. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction ?k? Express your answer using two significant figures. ?k = SubmitRequest Answer

Answer 1

(1) Initial energy of the spring mass system = spring energy = (1/2)kx²
where k is spring constant = 70 N/m
x is compression of the spring = 0.5 m
Initial energy = (1/2)*70*0.5² = 8.75 J
Final energy of the mass = potential energy of the mass = mgh
where h is height from the initial position
Now applying energy conservation
mgh = 8.75
h = 8.75/(9.81*2.2) = 0.405 m
Now distance along the slope
dSin = h
dSin41 = h
d = h/sin41 = 0.405/Sin41 = 0.618 m
Distance travelled along the slope will be 0.618 m
(b) When mass is attached to the spring then first mass will cover
0.5 m to reach equilibrium position then it will cover 0.5 m more to produce spring energy equal to the kinetic energy of the mass.
Therfore total distance travelled by the mass = 1 m
before getting to rest
(c)Intial energy of the system = 8.75 J ------------(a)
Now in distance 0.5 m the block reach at rest
Final energy of system = potential energy of block + frictional energy
= mg(0.5Sin41) + f_K*0.5
where f_k is friction force = u_k*R_N
Where R_N is normal reaction = mgCos41
f_k = u_k*(mgCos41)
Final energy = mg*(0.5Sin41) + u_K*(mgCos41)
= 2.2*9.81*(0.5Sin41) + u_k *(2.2*9.81*Cos41)
= 7.079 + 16.29u_k --------(b)
Now applying energy conservation
7.079 + 16.29u_k = 8.75
u_k = 0.102