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Zanzabum
2 years ago
5

An object experiences an acceleration of -6.8 m/s​2.​ As a result, it accelerates from 54 m/s to a complete stop. How much dista

nce did it travel during that acceleration?
Physics
1 answer:
inessss [21]2 years ago
3 0

Answer:

The distance traveled during its acceleration, d = 214.38 m

Explanation:

Given,

The object's acceleration, a = -6.8 m/s²

The initial speed of the object, u = 54 m/s

The final speed of the object, v = 0

The acceleration of the object is given by the formula,

                                      a = (v - u) / t   m/s²

       ∴                              t = (v - u) / a

                                         = (0 - 54) / (-6.8)

                                         = 7.94 s

The average velocity of the object,

                                       V = (54 + 0)/2

                                           = 27 m/s

The displacement of the object,

                                 d = V x t   meter

                                    = 27 x 7.94

                                    = 214.38 m

Hence, the distance the object traveled during that acceleration is, a = 214.38 m

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a) the impulse exerted by the rivet when the anvil has an infinite mass support is 0.932 lb.s

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b) the impulse exerted by the rivet when the anvil has a support weight of 9 lb = 0.799 lb.s

the energy absorbed by the rivet under each blow when the anvil has a support weight of 9 lb is = 7.99 ft.lb

Explanation:

The first picture shows a schematic view of a free body momentum diagram of the hammer head and the anvil.

Using the principle of conservation of momentum to determine the final velocity of anvil and hammer after the impact; we have:

m_Hv_H + m_Av_A = m_Hv_2+m_Av_2

From the question given, we can deduce that the anvil is at rest;

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m_Hv_H + 0 = (m_H+m_A) v_2

Making v_2 the subject of the formula; we have:

v_2 = \frac{m_Hv_H}{m_H + m_A}       ------- Equation  (1)

Also, from the second diagram; there is a representation of a free  body momentum  of the hammer head;

From the diagram;

F = impulsive force exerted on the  rivet

Δt = the change in time of application of the impulsive force

Using the principle of impulse of momentum to the hammer in the quest to determine the impulse exerted (i.e FΔt ) on the rivet; we have:

m_Hv_H - F \delta t = m_Hv_2

- F \delta t = - m_Hv_H + m_Hv_2

F \delta t = m_Hv_H - m_Hv_2

F \delta t = m_H(v_H - v_2)        ------- Equation   (2)

Using the function of the kinetic energy  of the hammer before impact T_1; we have:

T_1 = \frac{1}{2} m_Hv_H^2  -------- Equation (3)

We determine the mass of the hammer m_H  by using the formula from:

W_H = m_Hg

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W_H = weight of the hammer

m_H = mass of the hammer

g = acceleration due to gravity

Making m_H the subject of the formula; we have:

m_H = \frac{W_H}{g}

m_H = \frac{1.5 \ lb}{32.2 \ ft/s^2}

m_H = 0.04658 \ lb.s^2/ft

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T_1 = \frac{1}{2} m_Hv_H^2

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After the impact T_2 ; the final kinetic energy of the hammer and anvil can be written as:

T_2 = \frac{1}{2}(m_H +m_A)v^2_2

Recall from equation (1) ; where v_2 = (\frac{m_Hv_H}{m_H+m_A})  ; if we slot that into the above equation; we have:

T_2 = \frac{1}{2}(m_H +m_A)( \frac{m_Hv_H}{m_H+m_A})^2

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T_2 = \frac{1}{2} ({m^2_H +v^2})(\frac{m_H}{m_H+m_A})

Also; from equation (3)

T_1 = \frac{1}{2} m_Hv_H^2; Therefore;

T_2 = T_1 (\frac{m_H}{m_H+m_A})    ----- Equation (4)

a)

Now; To calculate the impulse exerted by the rivet FΔt and the energy absorbed by the rivet under each blow  ΔT when the anvil has an infinite mass support; we have the following process

First , we need to find the mass of the anvil when we have an infinite mass support;

mass of the anvil m_A = \frac{W_A}{g}

where we replace;  W_A \ with \ \infty and g = 32.2 ft/s²

m_A =  \frac{\infty}{32.2 \ ft/s}

However ; from equation (1)

v_2 = \frac{m_H v_H}{m_H + m_A}

v_2 = \frac{0.04658*20}{0.04658+ \ \infty}

v_2 = 0

From equation (2)

F \delta t = m_H(v_H + v_2)      

F \delta t = (0.04658 lb .s^2 /ft )(20ft/s  - 0)

F \delta t = \ 0.932 \  lb.s

Therefore the impulse exerted by the rivet when the anvil has an infinite mass support is  0.932 lb.s

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T_2 = T_1 (\frac{m_H}{m_H+m_A} )

where;

T_1= 9.316 \ ft.lb

m_H = 0.04658 \ lb.s^2/ft

m_A = \infty

Then;

T_2 = (9.316 \ ft.lb) (\frac{0.04658\  lb.s^2/ft)}{0.04658  \ lb.s^2/ft+ \infty} )

T_2 = (9.316 \ ft.lb)* 0

T_2 = 0

Then the energy absorbed by the rivet under each blow ΔT when the anvil has an infinite mass support

ΔT = T_1 - T_2

ΔT = 9.316 ft.lb - 0

ΔT ≅  9.32 ft.lb

Therefore; we conclude that the energy absorbed by the rivet under each blow  when the anvil has an infinite mass support = 9.32 ft.lb

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