Problem Set 6.8: | 1
| 3 | 4 | 5 | 6
| 17 | 19 | 21 | 23
| __25__ | 28
| 29 | 31 | Go up

** - by Sonja Scherer, Cooper Spear, and Chris Murray 2001**

**1. A 75.0-kg firefighter climbs a flight of stairs 10.0 m
high. How much work is required? **

Well, work is given by

W = Fs (if the force and the displacement are parallel, which I am going to assume)

The force needed to move the firefighter up the ladder is her weight, or

F = ma

F = (75 kg)(9.8 N/kg) = 735 N

And finally, W = Fs = (735 N)(10.0 m) = 7350 Nm or7350 J

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**3. How much work did the movers do (horizontally) pushing
a 150-kg crate 12.3 m across a rough floor without acceleration, if the
effective coefficient of friction was 0.70? **

Work is given by

W = Fs (if the force and the displacement are parallel, which I am going to assume)

The force would be the force of friction,

The normal force is the weight of the 150 kg crate or (150 kg)(9.80 N/kg) = 1470 N

So applying the friction formula: (Since it is not accelerating, the force needed is exactly the kinetic friction force)

F_{fr}= µ_{k}F_{N }F_{fr}= (.70)(1470 N) = 1029 N

And finally, W = Fs = (1029 N)(12.3 m) = 12656.7 Nm or1.3 x 10^{4}J

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**4. A car does 7.0x10 ^{4} J of work in traveling
2.8 km at a constant speed. What was the average retarding force acting on
the car?**

s = 2.8 km = 2800 m

Work is given by

W = Fs (if the force and the displacement are parallel, which I am going to assume)

so F = W/s = (7.0 x 10^{4}J)/(2800 m) =25 N

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**5. How high will a 0.325-kg rock go if thrown straight up
by someone who does 115 J of work on it? Neglect air resistance.**

Work is given by

W = Fs (if the force and the displacement are parallel, which I am going to assume)

The force needed to move the rock upward is its weight, or

F = ma

F = (.325 kg)(9.8 N/kg) = 3.185 N

And finally,

W = Fs, so

F = W/s = (115 J)/(3.185 N) =36.1 m

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**6. A hammerhead with a mass of 2.0 kg is allowed to fall
onto a nail from a height of .40 m . What is the maximum amount of work it could
do on the nail? Why do people not just "let it fall" but add their own
force to the hammer as it falls?**

The work it can do on the nail is the same as what was done to lift it, so

Work is given by

W = Fs (if the force and the displacement are parallel, which I am going to assume)

The force needed to lift the hammer is its weight, or

F = ma

F = (2.0 kg)(9.8 N/kg) = 19.6 N

And finally, W = Fs = (19.6 N)(.40 m) = 7.84 Nm or7.8 J

People push on the hammer to give it more energy. Since energy is the ability to do work, the more kinetic energy the hammer has when it strikes the nail, the more work it can do.

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**17. At room temperature, an oxygen molecule, with mass of
5.31 x 10 ^{-26} kg, typically has a KE of about 6.21 x 10 ^{-21} J. How fast
is it moving?**

The formula for kinetic energy is:

E_{k}=^{1}/_{2}mv^{2 }

We want the velocity, so let's solve for it first, as I don't really want to do math with numbers in scientific notation:

2E_{k}= mv^{2 }

2E_{k}/m = v^{2 }

v = Ö{2E_{k}/m} = Ö{2(6.21 x 10^{-21}J)/(5.31 x 10^{-26}kg)} = 483.6 m/s =484 m/s

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**19. How much work is required to stop an electron (m =
9.11 x 10 ^{-31} kg) which is moving with a speed of 1.90 x 10^{6} m/s**

You would need to do an amount of work equal to its kinetic energy. (Or rather, the electron would do work on you - giving you energy)

The formula for kinetic energy is:

E_{k}=^{1}/_{2}mv^{2 }

E_{k}=^{1}/_{2}(9.11 x 10^{-31}kg)(1.90 x 10^{6}m/s)^{2 }=^{ }1.64436x10^{-18}J =1.64x10^{-18}J

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**21. An automobile is traveling along a highway at 90
km/h. If it travels instead at 100 km/h, what is the percent increase in the
automobile's kinetic energy?**

We don't know the mass of the car, but don't let that get you down. We will just have to be clever.

The formula for kinetic energy is:

E_{k}=^{1}/_{2}mv^{2 }

now, we'll let v_{1}= 90 km/hr, and v_{2}= 100 km/hr

The percent increase is just one thing, divided by another times 100, so we have

%increase = E_{k1}/E_{k1 }= {^{1}/_{2}mv_{1}^{2}}/{^{}1/_{2}mv_{2}^{2}} - now the constants 1/2 and m cancel out:

%increase = E_{k1}/E_{k1 }= {v_{1}^{2}}/{v_{2}^{2}}_{ }= {(100 km/hr)^{2}}/{(90 km/hr)^{2}} = 1.23 or23% increase

The units don't matter, as they cancel out too. Life is good.

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**23. A baseball (m = 140 g) traveling 35 m/s moves a
fielder's glove backward 25 cm when the ball is caught. What was the average
force exerted by the ball on the glove?**

m = 140g = .140 kg, s = 25 cm = .25 m

This is a conservation of energy problem. The idea is that the kinetic energy of the moving baseball goes into the work it takes to move the glove backward. So I will set the kinetic energy equal to work:

^{1}/_{2}mv^{2 }= Fs

^{1}/_{2}(.140 kg)(35 m/s)^{2 }= F(.25 m)

F =343 N

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**25. At an accident scene on a level road, investigators
measure a car's skid mark to be 88 m long. It was a rainy day and the
coefficient of friction was estimated to be 0.42. Use these data to determine
the speed of the car when the driver slammed on (and locked) the brakes. (Why
does the car/s mass not matter?)**

Here the kinetic energy of the car goes entirely to work done against friction. (i.e. the tires sliding along the ground) I will set the kinetic energy equal to work:

^{1}/_{2}mv^{2 }= Fs

Where F is the force of friction given by F_{fr}= µ_{k}F_{N }- so I will substitute this for the force:

^{1}/_{2}mv^{2 }= Fs = (µ_{k}F_{N})s

And finally, the normal force, F_{N}is simply the weight in this case. F = ma, so weight equals mg where g is the acceleration of gravity. Substituting this in for the normal force:

^{1}/_{2}mv^{2 }= ( µ_{k}mg)s

Notice that at this point I can cancel the mass of the car from both sides, leaving:

^{1}/_{2}v^{2 }= ( µ_{k}g)s

Now we can solve for v:

v^{2 }= 2(µ_{k}g)s

v = Ö{2( µ_{k}g)s} = 26.9 m/s =27 m/s

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**28. A 220-kg load is lifted 21.0 m
vertically with an acceleration a = 0.150 g by a single cable. Determine (a) the
tension in the cable, (b) the net work done on the load, (c) the work done by
the cable on the load, (d) the work done by gravity on the load, and (e) the
final speed of the load assuming that it started from rest.**

a = .150 g = (.150)(9.8 m/s/s) = 1.47 m/s/s

a)

To calculate the tension in the cable, let's do a little Newton's second law here. We have the force of gravity downward of (220 kg)(9.8 N/kg) = 2156 N downward (-), and the cable tension T upward (+) and an upward acceleration of 1.47 m/s/s upward (+), so our F = ma look like:

<T - 2156 N> = (220 kg)(+1.47 m/s/s)

T = 2479.4 N

c)

The work done by the cable on the load is W = F^{.}s = (2479.4 N)(21.0 m) = 52067.4 J. This is a positive quantity as the displacement is in the direction of the force. (The cable pulled up, and it moved up)

d)

The work done by gravity is W = F^{.}s = (2156 N)(21.0 m) = 45276 J. This is a negative quantity because the displacement is up, but the force of gravity is down.

b)

The Net work done on the load is the sum of the work done by the cable on the load, and the work done by gravity on the load.

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**29. A spring has a spring constant, k, of 440 N/m. How
much must this spring be stretched to store 25 J of potential energy**

Well, here's really hoping that you don't need to actually consult this website to solve this problem.

E_{elas}=^{1}/_{2}kx^{2}, E_{elas}= 25 J, and k = 440 N/m, so x = .337 m

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**31. By how much does the gravitational potential energy
of a 64-kg pole vaulter change if his center of mass rises about 4.0 m during
the jump?**

It's pronounced Poh-Wahl Vawlt. I think this is the formula:

DE_{p}= mgDh, m = 64 kg, h = 4.0 m, so DE_{p}= 2508.8 J or about 2500 J. (They store this energy by running and building up kinetic energy)

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