6.4. PHYSICS APPLICATIONS: WORK, FORCE, AND PRESSURE
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Of course, the formula W = F · d only applies when the force is constant while it is
exerted over the distance d. In Preview Activity 6.4, we explore one way that we can use a
definite integral to compute the total work accomplished when the force exerted varies.
Preview Activity 6.4. A bucket is being lifted from the bottom of a 50-foot deep well;
its weight (including the water), B, in pounds at a height h feet above the water is given
by the function B(h). When the bucket leaves the water, the bucket and water together
weigh B(0) = 20 pounds, and when the bucket reaches the top of the well, B(50) = 12
pounds. Assume that the bucket loses water at a constant rate (as a function of height, h)
throughout its journey from the bottom to the top of the well.
(a) Find a formula for B(h).
(b) Compute the value of the product B(5)△h, where △h = 2 feet. Include units on
your answer. Explain why this product represents the approximate work it took to
move the bucket of water from h = 5 to h = 7.
(c) Is the value in (b) an over- or under-estimate of the actual amount of work it took
to move the bucket from h = 5 to h = 7? Why?
(d) Compute the value of the product B(22)△h, where △h = 0.25 feet. Include units
on your answer. What is the meaning of the value you found?
(e) More generally, what does the quantity W slice = B(h)△h measure for a given value
of h and a small positive value of △h?
(f) Evaluate the definite integral
50
0
B(h) dh. What is the meaning of the value you
find? Why?
⊲⊳
Work
Because work is calculated by the rule W = F · d, whenever the force F is constant, it
follows that we can use a definite integral to compute the work accomplished by a varying
force. For example, suppose that in a setting similar to the problem posed in Preview
Activity 6.4, we have a bucket being lifted in a 50-foot well whose weight at height h is
given by B(h) = 12 + 8e −0.1h .
In contrast to the problem in the preview activity, this bucket is not leaking at a
constant rate; but because the weight of the bucket and water is not constant, we have
to use a definite integral to determine the total work that results from lifting the bucket.
Observe that at a height h above the water, the approximate work to move the bucket a
small distance △h is
W slice = B(h)△h = (12 + 8e
−0.1h )△h.
367
Of course, the formula W = F · d only applies when the force is constant while it is
exerted over the distance d. In Preview Activity 6.4, we explore one way that we can use a
definite integral to compute the total work accomplished when the force exerted varies.
Preview Activity 6.4. A bucket is being lifted from the bottom of a 50-foot deep well;
its weight (including the water), B, in pounds at a height h feet above the water is given
by the function B(h). When the bucket leaves the water, the bucket and water together
weigh B(0) = 20 pounds, and when the bucket reaches the top of the well, B(50) = 12
pounds. Assume that the bucket loses water at a constant rate (as a function of height, h)
throughout its journey from the bottom to the top of the well.
(a) Find a formula for B(h).
(b) Compute the value of the product B(5)△h, where △h = 2 feet. Include units on
your answer. Explain why this product represents the approximate work it took to
move the bucket of water from h = 5 to h = 7.
(c) Is the value in (b) an over- or under-estimate of the actual amount of work it took
to move the bucket from h = 5 to h = 7? Why?
(d) Compute the value of the product B(22)△h, where △h = 0.25 feet. Include units
on your answer. What is the meaning of the value you found?
(e) More generally, what does the quantity W slice = B(h)△h measure for a given value
of h and a small positive value of △h?
(f) Evaluate the definite integral
50
0
B(h) dh. What is the meaning of the value you
find? Why?
⊲⊳
Work
Because work is calculated by the rule W = F · d, whenever the force F is constant, it
follows that we can use a definite integral to compute the work accomplished by a varying
force. For example, suppose that in a setting similar to the problem posed in Preview
Activity 6.4, we have a bucket being lifted in a 50-foot well whose weight at height h is
given by B(h) = 12 + 8e −0.1h .
In contrast to the problem in the preview activity, this bucket is not leaking at a
constant rate; but because the weight of the bucket and water is not constant, we have
to use a definite integral to determine the total work that results from lifting the bucket.
Observe that at a height h above the water, the approximate work to move the bucket a
small distance △h is
W slice = B(h)△h = (12 + 8e
−0.1h )△h.