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.(A) 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(B) Now the incline has a coefficient of kinetic friction μk. 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 μk?

Answer 1

A) d = 0.596m

B) μ = 0.206

Step-by-step explanation:

A) The potential energy stored in a spring when it is compressed is given as;

U = (1/2)kx² - - - - - - (eq1)

Where

U is potential energy stored in spring

k is spring constant

x is compressed length of the spring

Let the height the mass moved before coming to rest be h.

Thus, the potential energy at this height is;

U = mgh - - - - - - (eq2)

Since the mass is connected to the spring, according to the principle of conservation of energy, initial potential energy of the spring is equal to the sum of the final potential energy in the spring and the potential energy of the mass. Thus, we have;

(1/2)K(x1)² = (1/2)K(x2)² + mgh

Where x1 is the initial compressed length and x2 is the final compressed length.

Now, h will be dsin41 while x2 will be d - x1

Where d is the distance the mass moves up before coming to rest

Thus, we now have;

(1/2)K(x1)² = (1/2)K(d - x1)² + mg(dsin41)

(1/2)K(x1)² = (1/2)K(d² - 2dx1 + (x1)²) + mg(dsin41)

(1/2)K(x1)² = (1/2)Kd² - Kdx1 + (1/2)K(x1)² + mg(dsin41)

(1/2)Kd² - Kdx1 + mg(dsin41) = 0

(1/2)Kd² + d[mg(sin41) - Kx1] = 0

From the question,

k = 70 N/m

m = 2.2 kg

x1 = 0.5m

g = 9.8 m/s²

Thus, plugging in these values, we now have;

(1/2)(70)d² + d[(2.2•9.8•0.6561) - (70•0.5)] = 0

35d² - 20.8545d = 0

35d² = 20.8545d

Divide both sides by d to get;

35d = 20.8545

d = 20.8545/35

d = 0.596m

B) Here we are looking for the coefficient of friction.

First of all, let's find the kinetic energy at the equilibrium position;

The potential energy of the spring;

P.E = (1/2)K(x1)² = (1/2)(70)(0.5)² = 8.75J

The energy that will cause the block to decelerate = mg(x1)sin41 = 2.2 x 9.8 x 0.5 x 0.6561 = 7.073

So,

Net KE = 8.75 – 7.073 = 1.677J

Now, for the block to stop moving at this equilibrium position, the work done by the frictional force must be equal to KE of 1.677J

Thus,

F_f(x1) = 1.677J

F_f = μmgcos 41

Where, μ is the coefficient of friction;

So, F_f = μ(2.2 x 9.8 x 0.7547)

Ff = 16.271μ

Thus,

16.271μ x 0.5 = 1.677J

8.136μ = 1.677

μ = 1.677/8.136

μ = 0.206

Answer 2

A) = 0.63 m

B)This is approximately 0.21

Step-by-step explanation:

Part A)

Now if spring is connected to the block then again we can use energy conservation

so we will have

`(1)/(2)kx^2 = mg(x + x')sin\theta + (1)/(2)kx'^2`

so we will have

`(1)/(2)(70)(0.5^2) = 2(9.81) (0.50 + x') sin41 + (1)/(2)(70)x'^2`

`8.75 = 6.43 + 12.87 x' + 35 x'^2`

`x' = 0.13 m`

so total distance moved upwards is

`L = 0.5 + 0.13 = 0.63 m`

Part B

Now the incline has a coefficient of kinetic friction μk. 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 μk

KE = 8.75 – 10.78 * sin 41

This is approximately 1.678 J. This is the kinetic energy at the equilibrium position. For the block to stop moving at this position, this must be equal to the work that is done by the friction force.

Ff = μ * 10.78 * cos 41

Work = 0.5 * μ * 10.71 * cos 41

μ * 10.78 * cos 41 = 8.75 – 10.78 * sin 41

μ = (10 – 8.82 * sin 41) ÷ 8.82 * cos 41

μ = 1.678/8.136

μ = 0.206

This is approximately 0.21