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3 years ago

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A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide freely along the smooth vertical guides BC and

DE. Determine the velocity of the block and plate just after impact. The coefficient of restitution between the block and the plate is e = 0.75. Also, find the maximum compression of the spring due to impact.The spring has an unstretched length of 600 mm
The spring with the 5 kg plate are stationary before impact at 450mm. The 600mm unstretch length of the spring is without the 5 kg plate

1 answer:

Liono4ka [1.6K]3 years ago

8 0

Answer:

Check the explanation

Explanation:

<u>Given data </u>

Mass of the block A is Ma = 10kg

Mass of the plate is Mp = 5kg

Coefficient of restitution between the block and that of the plate is e=0.75

Unstretched length of the spring is 0.6m

From the conservation of energy, we have four motion of block A from (1) to (2)

$(Ta)_1 + (Va)_1 = (Ta)_2 + (Va)_2$

Then accordingly, we will have the step by step calculation in the attached image below .

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A four-cylinder, four-stroke internal combustion engine has a bore of 3.7 in. and a stroke of 3.4 in. The clearance volume is 16

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Answer:

the net work per cycle $\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, W = 88.0144746 hp

Explanation:

the information given includes;

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length of the stroke = 3.4 in

The clearance volume = 16% = 0.16

The cylindrical volume $V_2 = 0.16 V_1$

the crankshaft N rotates at a speed of 2400 RPM.

At the beginning of the compression , temperature $T_1$ = 60 F = 519.67 R

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$0.84 V_1 =36.55714291$

$V_1 =\dfrac{36.55714291}{0.84 }$

$V_1 =43.52040823 \ in^3 \\ \\ V_1 = 43.52 \ in^3$

$V_1 = 0.02518 \ ft^3$

the mass in air ( lb) can be determined by using the formula:

$m = \dfrac{P_1V_1}{RT}$

where;

R = 53.3533 ft.lbf/lb.R°

$m = \dfrac{2088 \ lb/ft^2 \times 0.02518 \ ft^3}{53.3533 \ ft .lbf/lb.^0R \times 519 .67 ^0 R}$

m = 0.0018962 lb

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At state of volume 2; the relative volume can be determined as:

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$W_{net} = Heat \ supplied - Heat \ rejected$

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$W_{net} = 0.0018962 \times (410.08)$

$\mathbf{W_{net} = 0.777593696}$ Btu per cycle

the power developed by the engine, in horsepower. can be calculated as follows;

In the four-cylinder, four-stroke internal combustion engine; the power developed by the engine can be calculated by using the expression:

$W = 4 \times N' \times W_{net$

where ;

$N' = \dfrac{2400}{2}$

N' = 1200 cycles/min

N' = 1200 cycles/60 seconds

N' = 20 cycles/sec

W = 4 × 20 cycles/sec × 0.777593696

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