Question

A block of mass m = 2.5 kg is attached to a spring with spring constant k = 730 N/m. It is initially at rest on an inclined plane that is at an angle of θ = 21° with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is μk = 0.19. In the initial position, where the spring is compressed by a distance of d = 0.11 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.
A) What is the block's initial mechanical energy?
B) 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

Answer:

A) Em = 4.41 J

B) L = 0.33m

Explanation:

A) The total mechanical energy of the block is the elastic potential energy due to the compressed spring. The gravitational energy is zero. Then you have:

$E_m=\frac{1}{2}k(\Delta x)^2$

k: constant's spring = 730 N/m

Δx: distance of the compression = 0.11m

You replace the values of k and Δx:

$E_m=\frac{1}{2}(730N/m)(0.11m)^2=4.41\ J$

B) To find the distance L traveled by the block you take into account that the total mechanical energy of the block is countered by the work done by the friction force, and also by the work done by the gravitational energy.

Then, you have:

$E_m-W_f-W_g=0\\\\W_f=(\mu Mg cos\theta)L\\\\W_g=(Mgsin\theta)L$

μ:  coefficient of kinetic friction = 0.19

g: gravitational acceleration = 9.8m/s^2

M: mass of the block = 2.5kg

θ: angle of the inclined plane = 21°

You replace the values of all parameters:

$E_m-W_f-W_g=0\\\\4.41-(0.19)(2.5kg)(9.8m/s^2)(cos21\°)L-(2.5kg)(9.8m/s^2)(sin21\°)L=0\\\\4.41-4.34L-8.78L=0\\\\4.41-13.12L=0\\\\L=0.33m$

hence, the distance L in which the block stops is 0.33m

Answer 2

(a) The block's initial mechanical energy on the given position is 4.42 J.

(b) The distance traveled by the block when it is pushed by the spring before coming to rest is 1.02 m.

The given parameters;

• mass of the block, m = 2.5 kg
• spring constant, k = 730 N/m
• angle of inclination, θ = 21°
• coefficient of friction, μ = 0.19
• compression of the spring, x = 0.11 m

The block's initial mechanical energy is calculated as follows;

$E = K.E _i + P.E_i\\\\E = \frac{1}{2} mv^2 \ + \ \frac{1}{2} kx^2\\\\E = \frac{1}{2} m (0)^2 \ + \ \frac{1}{2} \times 730 \times (0.11)^2\\\\E = 4.42 \ J$

The block will travel up if the energy applied by the spring is greater than the work-done by frictional force on the block.

The work-done on the block by the frictional force is calculated as follows;

$W_f = F_k \times d\\\\W_f= \mu_k F_n \times d\\\\W_f = \mu_k mgcos(\theta) \times d\\\\W_f = (0.19)(2.5)(9.8)cos(21) \times d \\\\W_f = 4.346 d$

Apply work-energy theorem;

$4.346d = 4.42\\\\d = \frac{4.42}{4.346} = 1.02 \ m$

Thus, the distance traveled by the block when it is pushed by the spring before coming to rest is 1.02 m.

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