Sample problem page

Main Page > User:Cmurray > Murray's Sand Box > Sample problem page == Problems ==

1. What force is needed to accelerate a child on a sled (total mass = 60.0 kg) at 1.15 m/s/s?

Assuming there is no friction, the any force we exert horizontally will be unbalanced, and so:

F = ma

F = (60.0 kg)(1.15 m/s/s) = 69 N

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2.

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3. How much force is required to accelerate a 9.0-g object at 10,000 "g's" (say in a centrifuge)?

They give us m = 9.0 g = 9.0 x 10^-3 kg, and a = 10,000 "g's" = 98000 m/s/s so:

F = ma = (9.0 x 10-3 kg)(98000 m/s/s) = 882 N = 880 N

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4.

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5. What is the weight of a 66-kg astronaut (a) on Earth, (b) on the moon (g = 1.7 m/s²), (c) on Mars (g = 3.7 m/s²), (d) in outer space traveling at constant velocity?

Weight (the force of gravity that causes the gravitational acceleration must equal ma because of Newton's second law:

on earth F = ma = (66 kg)(9.80 m/s/s) = 646.8 N = 650 N

on moon F = ma = (66 kg)(1.7 m/s/s) = 112.2 N = 110 N

on mars F = ma = (66 kg)(3.7 m/s/s) = 244.2 N = 240 N

At a constant velocity in space: F = ma = (66 kg)(0) = 0 N = 0 N

(Since the velocity is constant, there is no acceleration so they are weightless)

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6.

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7. What average force is required to stop an 1100-kg car in 8.0 s if it is traveling at 90 km/h?

First, notice that the initial velocity is given in km/hr, so let's first change that to m/s by dividing by 3.6:

90 km/hr/3.6 = 25 m/s

We have a cute linear kinematics problem to solve:

s = don't care

u = 90 km/hr/3.6 = 25 m/s

v = 0

a = ???

t = 8.0 s

Use v = u + at to find a:

v = u + at

0 = 25 m/s + a(8.0 s)

a = -3.125 m/s/s

so now we can use Newton's second law

F = ma = (1100 kg)(-3.125 m/s/s) = -3437.5 N = -3400 N

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8.

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9. A fisherman yanks a fish out of the water with an acceleration of 4.5 m/s² using very light fishing line that has a "test" value of 22 N. The fisherman unfortunately loses the fish as the line snaps. What can you say about the mass of the fish?

The weight of the fish is F = ma = mg down (-) and there is a possible force of 22 N up (+), so our expression of Newton's second law looks like this (up is +) The acceleration of 4.5 m/s/s is also positive because it is up:

<22 N - mg> = m(+4.5 m/s/s)

<22 - 9.8m> = m(4.5)

22 = 9.8m + 4.5m = 14.3m

m = 22/14.3 = 1.538 kg = 1.5 kg

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11. What is the average force exerted by a shot-putter on 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m/s?

We have a cute linear kinematics problem to solve first:

s = 2.8 m

u = 0 (assumed)

v = 13 m/s

a = ???

t = Don't care

Use v2 = u2 + 2as, a = 30.1786 m/s/s

F = ma = (7.0 kg)(30.1786 m/s/s) = 211.25 = 210 N

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13. A 10-kg bucket is lowered by a rope in which there is 63 N of tension. What is the acceleration of the bucket? Is it up or down?

First calculate the weight:

F = ma = (10 kg)(9.80 m/s/s) = 98 N

Now you have a force of 98 N down (-) and a force of 63 N up (+) and our F = ma looks like this if we make up positive:

<63 N - 98 N> = (10 kg) a

a = -3.5 m/s/s (down)

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14.

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15. A 75-kg petty thief wants to escape from a third-story jail window, Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 58 kg. How might the thief use this "rope" to escape? Give quantitative answer.

If it can support a mass of 58 kg, the breaking force is the weight of 58 kg or (58 kg)(9.80 N/kg) = 568.4 N. This would be the maximum upward force the sheets could supply.

Mr. thief could accelerate downward at a constant rate, thereby reducing the tension in the sheets to this value. So the thief would have 568.4 N of force upward (+), and his own weight (75 kg)(9.80N/kg) = 735 N down (-) and our F = ma looks like:

<568.4 N - 735 N> = (75 kg)a

a = -2.2213 m/s/s = -2.2 m/s/s. By accelerating downwards at 2.2 m/s/s

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16.

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17. The cable supporting a 2100-kg elevator has a maximum strength of 21,750 N. What maximum upward acceleration can it gave the elevator without breaking?

First calculate the weight:

F = ma = (2100 kg)(9.80 m/s/s) = 20580 N

Now you have a force of 20580 N down (-) and a force of 21750 N up (+) and our F = ma looks like this if we make up positive:

<21750 N - 20580 N> = (2100 kg) a

a = +.557 m/s/s = +.56 m/s/s upward

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