Problem Set 4.6: | 1 | 3 | 5 | 7 | 9 | 11 | 13 | 15 | 17  | Go up
- by Cooper Spear, Chris Murray, 2001

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

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

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)

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

We have a cute linear kinematics problem to solve first:
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

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> = ma
22 = 9.8m + 4.5m = 14.3m
m = 22/14.3 = 1.538 kg = 1.5 kg

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

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)

15. A 75-kg petty thief wants to escape form 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