(2012? Fengtai District, the second model) The figure is a science and technology team designed a pulley block model device. The pulley block is powered by an electric motor, and in the laboratory Min

(2012? Fengtai District, the second model) The figure is a science and technology team designed a pulley block model device. The pulley block is powered by an electric motor, and in the laboratory Ming and (1) in 0 ~ 7s, the metal block A to 0.1m / s speed in the liquid at a uniform rate of vertical rise, from the figure can be known as the power of the motor P1 = 6W;

∴F1=

< tr>

P1

V Rope

=

P1

n×V object

=

6W

3×0.1m/s

=20N

Answer: The tension F1 provided by the motor before the metal block is exposed to the liquid is 20N.

(2)①The metal block A rises and the liquid surface falls at the same time, and the height of A is equal to the distance A moves when the metal block leaves the liquid surface plus the height at which the liquid surface falls;

∴hA=VAt+

SA×VAt

S cylinder?SA

=0.1m/s×2s+

100cm2×0.1m/s × 2s

500cm2?100cm2

=0.25m

②The volume of the metal block A, VA=SA×hA=100×10-4m2× 0.25m=2.5×10-3m3

Gravity of metal block A GA=mAg=ρA×VA×g=3×103kg/m3×2.5×10-3m3×10N/kg=75N

③The total work done by the electric motor during the ascent of A when it is completely submerged in the liquid: WTotal=Pt=6W×5s=30J;

When A is completely submerged in the liquid, the total work done by the electric motor is: WTotal=6W×5s=30J. ∵ the mechanical efficiency of the pulley block when the metal block A rises submerged in the liquid is

7

12

;

∴The useful work of this process W useful=W total η=30J x

7

12

=17.5J

And because of this process the object A rises a distance of h=0.1m/s × 5s=0.5m

< p>Then the useful work done W useful = (GA - F floating) h = (GA - ρ liquid gVA) h = (75N - ρ liquid × 10N/kg × 2.5 × 10-3m3) × 0.5m = 17.5J

Solve: ρ liquid = 1.6 × 103kg/m3?

④ Before and after the metal block A leaves the liquid surface, the height of the liquid level in the glass cylinder falls △h=< table cellpadding="-1" cellspacing="-1" style="margin-right:1px">VAS cylinder=

2.5×10?3m3

500×10?4m2

=0.05m

So the change in pressure of the liquid on the bottom of the container: P=ρliquid g△h=ρliquid ghA=1.6×103kg/m3×10N/kg× 0.05m=800Pa.

Answer: metal block A from the upper surface contact with the liquid surface to the lower surface to leave a *** took 2 seconds, before and after leaving the liquid surface, the liquid on the bottom of the container of the amount of change in the amount of pressure 800Pa.