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valentinak56 [21]

3 years ago

15

A block pushed along the floor with velocity v0xv0x slides a distance dd after the pushing force is removed. If the mass of the

block is doubled but its initial velocity is not changed, what distance does the block slide before stopping?

1 answer:

Volgvan3 years ago

5 0

Answer:

There is no change in distance

Explanation:

Given that block slides a distance dd with initial velocity $v_{ox}$ $v_{ox}$

Work energy theorem states that Work done W is the difference between final kinetic energy $KE_{f}$ and initial kinetic energy $KE_{i}$

W = $KE_{f}$ - $KE_{i}$ ...........(equation 1)

In the given problem, the work is done by frictional force

Hence the work done is given by

$W_{Friction}$ = - $F_{Friction}$ x dd

$F_{Friction}$ is negative since it is resistive force and dd is the distance

$W_{Friction}$ = - μ mg X dd

where μ is coefficient of friction.

Also we know that Kinetic energy KE = $\frac{1}{2}$ $mv^{2}$

$KE_{f}$ = 0 since final velocity is zero

$KE_{i}$ = $\frac{1}{2}$ $m(v_{ox} v_{ox}) ^{2}$

Substituting the corresponding values equation 1 becomes

-μ mg x dd = 0 - $\frac{1}{2}$ $m(v_{ox} v_{ox}) ^{2}$

dd= $\frac{(v_{ox}v_{ox} )^{2} }{2g}$ x (1/μ) ........... (equation 2)

To find distance when mass of block is doubled and initial velocity is not changed:

Equation 2 shows that the distance dd is independent of mass, therefore there is no change in distance.

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tatuchka [14]

Answer:

The question is incomplete as some details are missing, here is the details ;

In this final state

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Explanation:

From the concept of electrostatics, if a positively charged sphere is brought close to #2, there will be attraction of the opposite charges(-ve) towards it.

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3 years ago

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olga55 [171]

Answer:

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Explanation:

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3 years ago

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WITCHER [35]

The question is poor.

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8 0

4 years ago

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castortr0y [4]

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4 years ago

A piston–cylinder device with a set of stops initially contains 0.6 kg of steam at 1.0 MPa and 400°C. The location of the stops

Ilya [14]

Answer:

(a) Compression work at the final state with a pressure of 1(MPa) is: 44.32(KJ), (b) Compression work at the final state with a pressure of 500(KPa): 110.37(KJ) and (c) temperaure of the final state in part b: T=151.83(°C).

Explanation:

Remember that the substance is steam so it's water (H2O) and the initial conditions are $P_{1} =1MPa, T_{1}=400^{0}C, m=0.6Kg$ and $v_{2} =0.4v_{1}$ from a saturated water table and the initial conditions we can determine that the state phase is superheated (see Table 1 attached) because the $T_{sat}=179.88^{0} C \leq T_{1}$ from the table 1 we get: $v_{1} =0.30661(m^{3}/Kg)$ . Now we have second conditions as: $P_{2}=1(MPa), T_{2}=250^{0}C$ so from the same table we can see the state still superheated and we get $v_{2}=0.23275(m^{3}/Kg)$ , knowing that it's a isobaric process we can find the compression's work as: $W_{b}=m*P(v_{2}-v_{1})=0.6*1000*(0.23275-0.30661)=-44.32(KJ)$ so the compressor's work is: 44.32(KJ). (b) Then the piston reaches the stop and there are two processes in this stage, so Process 1 is isobaric and: $W_{1}=m*P*(v_{2}-v_{1}) =0.6*1000*(0.4*0.30661-0.30661)=-110.38(KJ)$ and the second process is isochoric: $W_{2}=zero$ ,now $W_{b}=W_{1}+ W_{2} =110.38+0=110.38(KJ)$ . Finally to get the temperarure at the final state in part (b) we get: $v_{2} =0.4v_{1} =0.4*0.30661=0.122644(m^{3}/Kg), P_{2}=500(KPa)$ from table 2 (see attached) we compare $v_{f}$ and $v_{g}$ at the saturated water table and find the following: $v_{f}=0.001093(m^{3}/Kg)$ , so we know that the final state phase is a satured mixture and we get the temperature at the final state as: $T_{2} =T_{sat} =151.83^{0}C$ .

3 0

4 years ago