Physics, 16.10.2020 06:01 sashllely6112

A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (Figure a). The object has a speed of vi = 2.60 m/s when it makes contact with a light spring (Figure b) that has a force constant of 50.0 N/m. The object comes to rest after the spring has been compressed a distance d (Figure c). The object is then forced toward the left by the spring (Figure d) and continues to move in that direction beyond the spring's unstretched position. Finally, the object comes to rest a distance D to the left of the unstretched spring (Figure e). The right end of a horizontal spring labeled k is attached to a wall. Five images show five configurations as a block labeled m approaches, compresses, and then moves away from the spring.
In figure a, the block is to the left of the spring, and an arrow above the block points to the right.
In figure b, the block is just touching the uncompressed spring, and an arrow labeled vector vi above the block points to the right.
In figure c, the block has compressed the spring by a distance d, and a label indicates vector vf = 0.
In figure d, the block is just touching the uncompressed spring, and an arrow labeled vector v above the block points to the left.
In figure e, the block is a distance D away from the spring, and a label indicates vector v = 0.
(a)
Find the distance of compression d (in m).
m
(b)
Find the speed v (in m/s) at the unstretched position when the object is moving to the left (Figure d).
m/s
(c)
Find the distance D (in m) where the object comes to rest.
m
(d)
What If? If the object becomes attached securely to the end of the spring when it makes contact, what is the new value of the distance D (in m) at which the object will come to rest after moving to the left?
m

A 1.10-kg object slides to the right on a surface having a coefficient of kinetic friction 0.250 (F

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Answer from: pearl24

The energy transferred between the object and the spring as a closed system, therefore, conserved are;

(a) The distance of compression, d ≈ 0.3354 meters

(b) The speed in the un-stretched position wen the object is sliding to the left, v ≈ 1.8623 m/s

(d) The distance the object will come to rest attached to the spring, D ≈ 0.2278 m

The reason the above values are correct are as follows;

The known parameters are;

Mass of the object, m₁ = 1.10 kg

Coefficient of friction, μ = 0.250

The initial speed of the object, v_i = 2.60 m/s

Force constant of the spring, K = 50.0 N/m

Distance the spring is compressed by the object = d

(a) Conservation of energy principle

$Kinetic \ energy = \dfrac{1}{2} \cdot m\cdot v^2$

Work done = Force × Distance

Friction force, F_f = W × μ

Weight, W = m·g

Weight = Mass × Acceleration

Energy transferred by object = Work done by spring + Work done by friction

$Energy \ transferred \ by \ object = Kinetic \ energy = \dfrac{1}{2} \times 1.10\times 2.60^2 = 3.718$

Energy transferred by object = 3.718 J

$Work \ done \ by \ spring = \dfrac{1}{2} \cdot k\cdot x^2$

$Work \ by \ spring \ to \ bring \ object \ to \ rest, \ W_{spring} = \dfrac{1}{2} \times 50\times d^2$

$W_{spring}$ = 25·d²

Work done by friction, $W_{friction}$ = 1.10×9.81×0.250×d = 2.69775·d

Therefore;

3.718 = 25·d² + 2.69775·d

25·d² + 2.69775·d - 3.718 = 0

Solving gives

The distance of the compression d ≈ 0.3354 m

(b) The energy given by the spring = 25·d²

The work done by friction, $W_{friction}$ = 2.69775·d

Kinetic energy given to object = 0.55·v²

0.55·v² = 25·d² - 2.69775·d

0.55·v² = 25×0.3354² - 2.69775×0.3354

∴ v = √(3.4682) = 1.8623

The velocity of the object at the un stretched position, v ≈ 1.8623 m/s

K.E. = 0.5 × 1.10 kg × 3.4682 m²/s² = 1.90751 J

The work done by friction before object comes to rest = 2.69775·D

$D = \dfrac{1.90751 \, J}{2.69775 \, N} \approx 0.7071 \, m$

The distance where the object comes to rest, D ≈ 0.7071 m

(d) The work done on spring, $W_{spring}$ = 25·D'²

Work done on friction, $W_{friction}$ = 2.69775·D'

Kinetic energy of object, K.E. ≈ 1.90751 J

K.E. = $W_{spring}$ + $W_{friction}$

1.90751 ≈ 25·D'² + 2.6775·D'

25·D'² + 2.6775·D' - 1.90751 = 0

Solving with a graphing calculator gives;

D' ≈ 0.2278 m

The new value of the distance D = 0.2278 m

Learn more about the energy conservation principle here:

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Answer from: kylee76

(a) Approximately $0.335\; \rm m$ .

(b) Approximately $1.86\; \rm m\cdot s^{-1}$ .

(d) Approximately $0.228\; \rm m$ .

Explanation:

denotes the velocity of the object in the first diagram right before it came into contact with the spring. Let

denote the mass of the block. Let $\mu$ denote the constant of kinetic friction between the object and the surface. Let

denote the constant of gravitational acceleration.Let

denote the spring constant of this spring.(a)

Consider the conversion of energy in this object-spring system.

First diagram: Right before the object came into contact with the spring, the object carries kinetic energy $\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^2$ .

Second diagram: As the object moves towards the position in the third diagram, the spring gains elastic potential energy. At the same time, the object loses energy due to friction.

Third diagram: After the velocity of the object becomes zero, it has moved a distance of and compressed the spring by the same distance.

Energy lost to friction: $\underbrace{(\mu \cdot m \cdot g)}_{\text{friction}} \cdot D$ . Elastic potential energy that the spring has gained: $\displaystyle \frac{1}{2}\,k\, D^2$ .

The sum of these two energies should match the initial kinetic energy of the object (before it comes into contact with the spring.) That is:

$\displaystyle \frac{1}{2}\, m \cdot {v_{i}}^{2} = (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2$ .

Assume that $g = 9.81\; \rm m \cdot s^{-2}$ . In the equation above, all symbols other than have known values:

$m =1.10\; \rm kg$ . $v_i = 2.60\; \rm m \cdot s^{-1}$ . $\mu = 0.250$ . $g = 9.81\; \rm m \cdot s^{-2}$ . $k = 50.0\; \rm N \cdot m^{-1}$ .

Substitute in the known values to obtain an equation for (where the unit of $D\!$ is .)

$3.178 = 2.69775\, D + 25\, D^2$ .

$2.69775\, D + 25\, D^2 + 3.178 = 0$ .

Simplify and solve for . Note that D 0 because the energy lost to friction should be greater than zero.

$D \approx 0.335\; \rm m$ .

(b)

The energy of the object-spring system in the third diagram is the same as the elastic potential energy of the spring:

$\displaystyle \frac{1}{2}\,k\, D^2 \approx 2.81\; \rm J$ .

As the object moves to the left, part of that energy will be lost to friction:

$(\mu \cdot m \cdot g) \, D \approx 0.905\; \rm J$ .

The rest will become the kinetic energy of that block by the time the block reaches the position in the fourth diagram:

$2.81\; \rm J - 0.905\; \rm J \approx 1.91\; \rm J$ .

Calculate the velocity corresponding to that kinetic energy:

$\displaystyle v =\sqrt{\frac{2\, (\text{Kinetic Energy})}{m}} \approx 1.86\; \rm m \cdot s^{-1}$ .

(c)

As the object moves from the position in the fourth diagram to the position in the fifth, all its kinetic energy ( $1.91\; \rm J$ ) would be lost to friction.

How far would the object need to move on the surface to lose that much energy to friction? Again, the size of the friction force is $\mu \cdot m \cdot g$ .

$\displaystyle (\text{Distance Travelled}) = \frac{\text{(Work Done by friction)}}{\text{(Size of the Friction Force)}} \approx0.707\; \rm m$ .

(d)

Similar to (a), solving (d) involves another quadratic equation about .

Left-hand side of the equation: kinetic energy of the object (as in the fourth diagram,) $1.91\; \rm J$ .

Right-hand side of the equation: energy lost to friction, plus the gain in the elastic potential energy of the spring.

$\displaystyle {1.91\; \rm J} \approx (\mu\cdot m \cdot g) \cdot D + \frac{1}{2}\, k \cdot D^2$ .

$25\, D^2 + 2.69775\, D - 1.90811\approx 0$ .

Again, D 0 because the energy lost to friction is greater than zero.

$D \approx 0.228\; \rm m$ .

Answer from: Quest

0 = 0.40 / 0 = 0

= 0.20 = 9.80 ⁄2 =?

=0 , = = 0.20 =0.50 0 0.40 ⁄

=0−122 =0−12(9.80 ⁄2)(0.50)2 =1.2

Answer from: Quest

The boat's velocity relative to the riverbank can be resolved into two components. the component going north is 2.7 m/s and the component going west is 1.2 m/s.we add these two vectors and then find the magnitude of the resultant to get the speed.the magnitude of the resultant vector from adding can be obtained by the pythagorean theorem. $\begin{aligned} | \vec{v}| & = \sqrt{(\text{2.7 m/s})^2 + (\text{1.2 m/s})^2 } \\ & = 2.954 \text{ m/s} \end{aligned}$ to two significant figures, the boat's speed relative to the riverbank is 3.0 m/s.

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