1.An aquarium 4 m long, 1 m wide, and 1 m deep is full of water. Find the work n

Question

1.An aquarium 4 m long, 1 m wide, and 1 m deep is full of water. Find the work needed to pump half of the water out of the aquarium. (Use 9.8 m/s2 for g and the fact that the density of water is 1000 kg/m3.)
Show how to approximate the required work by a Riemann sum. (Enter xi* as xi.)

2. A cable that weighs 6 lb/ft is used to lift 700 lb of coal up a mine shaft 550 ft deep. Find the work done.
Show how to approximate the required work by a Riemann sum. (Enter xi* as xi.)

3. A spring has natural length 26 cm. Compare the work W1 done in stretching the spring from 26 cm to 36 cm with the work W2done in stretching it from 36 to 46 cm. (Use k for the spring constant)

4. Suppose that 6 J of work is needed to stretch a spring from its natural length of 32 cm to a length of 46 cm.

(a) How much work is needed to stretch the spring from 36 cm to 38 cm? (Round your answer to two decimal places.)

(b) How far beyond its natural length will a force of 10 N keep the spring stretched? (Round your answer one decimal place.)

5. A spring has a natural length of 20 cm. If a 24-N force is required to keep it stretched to a length of 24 cm, how much work W is required to stretch it from 20 cm to 22 cm? (Round your answer to two decimal places.)

6. A force of 20 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching it from its natural length to 11 in. beyond its natural length?

7. variable force of 3x?2pounds moves an object along a straight line when it is x feet from the origin. Calculate the work done in moving the object from x = 1ft to x = 14ft. (Round your answer to two decimal places.)

Please show work

Explanation / Answer

1)

I assume the water is to be pumped to the height of the top of the aquarium.

The first thing to notice is that you don't need to use calculus. We can assume the water is of constant density, so the work required will be equal to the work required to raise a point mass equal to the mass of the water to be pumped and located at its centre of gravity.

The water to be pumped has an elevation of 0.5 to 1 metre.
Thus the centre of gravity of the water to be pumped has an elevation of 0.75 metres, so the point mass has to be raised through a height of 0.25 metres.

Work done = potential energy gained = mgh

where m = mass, g = acceleration due to gravity and h = height

Mass = volume * density
Thus the mass of the water = 4 m * 1 m * 0.5 m * 1,000 kg/m^3
= 2,000 kg

g = 9.8 m/s^2

So the work required to pump half of the water out of the aquarium
= 2,000 kg * 9.8 m/s^2 * 0.25 m
= 4,900 Joules

2)

Let x = the distance, in feet, of the coal from the top of the mine shaft.

At any height x, the total weight w, in pounds, of the cable/coal system is the weight of the coal (700 pounds) plus two pounds per every foot of cable that is hanging down:

w(x) = 700 + 2x

The amount of work done is the total displacement of this system, evaluated from x = 0 to x = 550:

?w dx = ?700 + 2x dx

= 700x + x

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.