Preparation problems

7-6. A meteorite plunges to Earth, embedding itself 75 cm in the ground. If it does 140 MJ of work in the process, what average force does the meteorite exert on the ground?

Solution:

We can start out by using our definition of work:

W = ó
õ 		r2

r1 
 ®
F
 
 ·d ®
r
 
(1)
We can assume that the meteorite traveled in a straight line, making this a one-dimensional problem. Since the force the meteorite exerts on the ground is in the direction of travel, we don't have to worry about the dot product either. So Equation (1) becomes
W = ó
õ 		x2

x1 
 F  dx
Here we are given the total work and a path length and asked to find the average force exerted by the meteorite on the ground. Thus, we don't have to worry about doing the integral, as the average force will be constant over the entire path length. So we can simplify further to
W = Fav  Dx
Now we see that
Fav = W
Dx
Putting in our known values, we get
Fav
=
W
Dx
=
140 ×106  J
0.75  m
Fav
=
1.86 ×108  N

7-8. Two people push a stalled car at its front doors, each applying a 330-N force at 25° to the forward direction, as shown in Fig. 7-24 (on p. 171). How much work does each do in pushing the car 6.2 m?

Solution:

We can start out again by using our definition of work:

W = ó
õ 		r2

r1 
 ®
F
 
 ·d ®
r
 
(1)

Each person provides a constant force which is in a direction different than the motion of the car. The constant force allows us to dispense of the integral, but we must keep the dot product. So Equation (1) becomes

W = ®
F
 
 ·D ®
r
 
Since the force vector makes an angle of 25° with the direction of travel, we can use the dot product relationship from our constants and equation sheet
®
A
 
 · ®
B
 
 = | A || B | cosq
This gives us
W = | F || Dx | cosq
Putting in our known values, we get
W
=
(330  N)(6.2  m) cos(25°)
W
=
1854  J

7-22. Find the work done by a force F = 2.5 i + 3.7 j N as it acts on an object moving from the origin to the point r = 6.1 i + 2.4 j m.

Solution:

We start out again by using our definition of work:

W = ó
õ 		r2

r1 
 ®
F
 
 ·d ®
r
 
In this problem our force and direction of displacement are both constant, meaning we can drop the integral, giving us
W = ®
F
 
 ·D ®
r
 
Here we can find the dot product simply by noting that the x-components of both the force and the displacement lie along the same direction, as do the y-components. The dot product should then be given by
®
A
 
 · ®
B
 
 = Ax Bx + Ay By
The work is then given by
W = Fx rx + Fy ry
Putting in our known values, we get
W
=
(2.5  N)(6.1  m) + (3.7  N)(2.4  m)
W
=
15.25  J + 8.88  J
W
=
24.13  J

7-28. Find the total work done by the force shown in Fig. 7-29, as the object on which it acts moves from x = 0 to x = 5 m.

Solution:

We start out again by using our definition of work:

W = ó
õ 		r2

r1 
 ®
F
 
 ·d ®
r
 
We are operating in one dimension, so we can drop the vector symbols, and we notice that what we get looks alot like the calculation of the area underneath a curve:
W = ó
õ 		5

0 
 F(x) dx
Since we don't know the equation for the force as a function of distance, we can find the work by determining the area underneath the curve. This we can do by counting the squares, and recalling the formulas for the areas of a triangle and trapezoid
A (triangle)
=
1/2 (base)(height)
A (trapezoid)
=
1/2 (base1+base2)(height)
The work is then given by
W
=
1/2(2  N)(1  m) + (2  N)(1  m) + 1/2[(2  N)+(1  N)](1  m)
+ 1/2[(1  N)+(4  N)](1  m) + 1/2[(4  N)+(3  N)](1  m)
W
=
10.5  J

Physics 110 Home Page Physics 110 Honors Home Page

Solutions translated from TEX by TTH, version 1.57.