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faltersainse [42]
3 years ago
14

A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a cons

tant force of 2650 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5 m/s.
Engineering
1 answer:
sammy [17]3 years ago
3 0

Answer:

  t = 0.113 s

Explanation:

In this exercise we can use Newton's second law

X Axis            F - fr = ma

Y Axis           N –W = 0

                    N = W = m g

The force of friction is

                  fr = μ N

We replace

                 F - μ m g = m a

                  a = (F- μ mg) / m

                 a = (2650 - 0.45 300 9.8) / 300

                 a = 4,423 m / s2

Now we can use the kinematics expressions, where we assume that the system starts with zero speed

                v = v₀ + a t

                t = v / a

                t = 0.5 / 4.423

                t = 0.113 s

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The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k
sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

8 0
2 years ago
Which option best describes a way engineers can provide maintenance, diagnoses, upgrades, or duplicates even for products they d
guajiro [1.7K]

Answer:

Reverse Engineering

Explanation:

Just took the test

4 0
2 years ago
Search and destroy christian
Setler [38]

Answer: ummmm wut?

Explanation:

3 0
3 years ago
Read 2 more answers
A lower coefficient of thermal conductivity along with a higher coefficient of thermal expansion and higher electrical resistanc
HACTEHA [7]

Answer:

stainless steel

Explanation:

Conductivity refers to the degree to which a specified material conducts electricity. It is the ratio of the current density in the material and the electric field.

The thermal conductivity of a material measures its ability to conduct heat.

In materials of low thermal conductivity, heat transfer occurs at a lower rate as compared to materials of high thermal conductivity.

Thermal expansion of the material refers to its tendency to change its shape, area, and volume as a result of change in temperature.

The electrical resistance of a material refers to the measure of its opposition to the flow of electric current.

<u>Stainless steel</u> has a lower coefficient of thermal conductivity along with a higher coefficient of thermal expansion and higher electrical resistance.

5 0
2 years ago
A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di
Bezzdna [24]

Answer :

<h3>Flow rate in pipe B is = 0.3094 \frac{m^{3} }{s}</h3>

Explanation:

Given :

Length of pipe A L_{A}  = 1500 m

Length of pipe B L_{B} = 2500 m

Flow rate through pipe A Q_{A}  = 0.4 \frac{m^{3} }{s}

Diameter of pipe D = 30 \times 10^{-2} m

Velocity from pipe A,

  V _{A} = \frac{Q_{A} }{A}

  V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2}  }

  V_{A}  = 5.66 \frac{m}{s}

Here, head loss is same because height is same.

    h_{a} = h_{b}

L_{A} V_{A} ^{2} = L_{B}  V_{B} ^{2}

V_{B} = \sqrt{\frac{1500}{2500}}    (5.66)

V_{B} = 4.38 \frac{m}{s}

Now rate of flow from pipe B is,

Q_{B}  = V_{B} A

Q_{B}  = \frac{\pi }{4}  (0.3)^{2} \times 4.38

Q_{B} = 0.3094 \frac{m^{3} }{s}

4 0
2 years ago
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