Q3: No, if the path is circular (which it will be if the Net force is centripetal), since the direction of the displacement (i.e., the motion) is always perpendicular to the centripetal force. Thus the scalar product between the force and the displacement is zero.

Q6: Yes,if the angle between the two vectors is greater than 90 degrees. Then its cosine will be negative, and thus the dot-product will be negative. Comment: there are always two angles between two vectors, $\theta \$ and 360 - $\theta \$. By convention, we use the smaller of the two, so that $\theta \$ is less than (or equal to) 180 degrees. If the angle is between 0 and 90 degrees, the cosine is (and thus the dot-product) is positive; if between 90 and 180 degrees, both are negative.

Q8. No! The dot product is also called the scalar product for the precise reason that it does NOT have direction. It is a simple number (like mass, for example, or time).

P9: In this case the force producing the motion and the displacement are in the same direction. We know that Work = Force x displacement.

We can get the force from F = ma = (6 kg)(2.0 m/s2) = 12 N

From Equation 2-12b on page 29, the displacement is $\Delta \$x = vot + 0.5 at2 = 0 + (0.5)(2.0)(72)

so $\Delta \$x = 49 m.

Thus

Work = (12 N)(49 m) = 588 Joules

P11: See Figure 5-8(b) and (c) on page 117 for the x-y coordinate system and how the angle $\theta \$ is defined.

We are told to ignore friction in this problem. Three forces act: (1) A normal force perpendicular to the incline surface (the +y direction), (2) gravity (i.e. the piano's weight), which acts downward, at an angle of 27 degrees CCW from the -y axis, and (3) the force exerted by the man, which is parallel to the incline surface and "up" the incline (along the -x axis).

We first must find this "applied force" (i.e., the force exerted by the man). Since the piano moves with constant speed (it is "kept from accelerating" as it slides down the incline), we can equate the x-component of the weight with the applied force:

FA = (Mg)sin( $\theta \$) = 1691 Newtons in the -x direction ("up") the incline

The displacement, meanwhile, is "down" the ramp in the +x direction.

(a) Work done by applied force = (FA)( $\Delta \$x)cos(180) = - 6594 Joules

(b) Work done by gravity = (Mg)( $\Delta \$x)cos(90 - $\theta \$ = + 6594 Joules

(c) The normal force does no work, since it is perpendicular to the displacement, thus WN = (FN)( $\Delta \$x)cos(90) = 0. Therefore the net work done (the sum of the work done by all three forces is

WNet = - 6594 + 6594 + 0 = 0 Joules

Comment: The net work must be zero if the piano's KE is not changing. We know this from the work-KE principle that in Section 7.4.

P16: We use the very important result of Equation 7-4 on page 168 (the proof is given just above the equation): A $\bullet \$B = AxBx + AyBy + AzBz

So, in this case:

A $\bullet \$B = (2)(11) + (-4)(2.5) + 95)(0) = 12.

P18: We have two ways of writing the scalar ("dot") product:

A $\bullet \$B = (A)(B)cos( $\theta \$)

A $\bullet \$B = AxBx + AyBy + AzBz

Equating these two expressions, we can solve this for cos( $\theta \$) and then use the inverse-cosine (or Arc-cosine) function on our calculator to obtain the angle $\theta \$. We first need:

A = sqrt(Ax2 + Ay2 + Az2) = 9.81

B = sqrt(Bx2 + By2 + Bz2) = 11.02

so

(9.81)(11.02)cos( $\theta \$) = (6.8)(8.2) + (-3.4)(2.3) + (-6.2)(-7.0)

with the result that cos( $\theta \$) = 0.844 $\theta \$ = 32.3 degrees

P21: We need a vector B = iBx + jBy such that

A $\bullet \$B = 0

Thus: AxBx + AyBy = 0

or

3Bx +1.5By = 0

So any vector B with the property that By = - 2Bx will do. For example, B = 2i - 4j will work. Verify this, and try it with another B of your own creating.

P40 Work is given by Equation 7-7 on page 169:

W = $\textstyle \int _{{a}}^{{b}}F\,dx$

Since we are given a graph of Force as a function of x, the integral is simply the area under the curve. So

(a) The area under F(x) over the interval (0, 10) can be easily computed by considering it as two triangles and a rectangle. The first triangle has base 3 m and height 400 N, so the area is (1/2)x(base)x(height) = 600 Joules. The rectangle over the region (3, 7) has base 4 m and height 400 N, so its area is 1600 J. Finally, right-hand triangle has the same dimensions as the first, so its area is also 600 J. The total area, and therefore the work, is 2800 Joules.

(b) The region (10, 12) appears to be a triangle, whose area is -200 Joules. The graph is poorly drawn beyond about x = 12, and it's not clear exactly where the flat line curves upward. I'll assume that the "corner" happens at x = 13.5 m. In this case, the area of the rectangular part is -300 J and that of the final triangle is -150 J. So the work from (10, 15) is -650 J. So, adding this to the answer in part (a), the total work from (0, 15) is 2150 Joules.

P56: A force of 105 Newtons acting over a distance of 0.75 meters does 79 Joules of work. If we neglect all other forces (see below), then this "net" amount of work must show up as the kinetic energy of the 0.085 kg arrow. Thus

(1/2)Mvf2 - (1/2)Mvo2 = 79

and, since the initial speed was zero, we have simply

(1/2)(0.085)vf2 = 79 which yields vf = 43.1 m/s

Can we justify neglecting other forces, especially gravity? If the arrow is fired horizontally (i.e., the 0.75 m is horizontal) then gravity does no work over this short accelerating distance, thus the "net" work is unaffected. If the arrow is fired vertically, then the gravitational force (Mg) does work equal to Mg $\Delta \$y = (0.085)(9.8)(0.75) = 0.62 Joules. Thus the net work differs from the 79 Joules above by this amount, which is less than one percent. For angles between horizontal and vertical, the gravitational work lies between these two bounds. So, we can neglect the work done by gravity and we will still be accurate to within 1%..

P63: Draw a free-body diagram for the block. Note that the displacement is "up" the incline, parallel to the incline surface.

(a) Work done by FP: The angle between FP and the displacement is 32 degrees. Thus

WP = (150)(5)cos(32) = 636 Joules

(b) Work done by gravity: The angle between the weight (Mg, directly downward) and the displacement is (32 + 90) = 122 degrees. Thus

WG = (18)(9.8)(5)cos(122) = - 467 Joules

(c) The angle between the normal force and the displacement is zero, so no work is done by the normal force.

(d) The net work is 636 - 467 = 169 Joules, which must be the amount of increase in the KE. Since the original KE is zero, we have

(1/2)Mvf2 = 169

and since M = 18 kg, the result is vf = 4.33 m/s

P79: Here the quickest (and, really, the only way using the math required for this course) is to take the scalar product F $\bullet \$d using Equation 7-4 on page 168. Doing this we get, for the work (note that Force is in kiloNewtons, so when we multiply by meters we will get work in kiloJoules):

W = Fxdx + Fydy + Fzdz

or

W = (10.0)(5.0) + (9.0)(4.0) + (12.0)(0) = 86.0 kiloJoules.

The angle between the two vectors is calculated using this result for the scalar product and also the fact that the scalar product must equal (F)(d)cos( $\theta \$). We compute the magnitudes F and d, then equate the two expressions, solving for cos( $\theta \$). Finally we use our calculator to find the inverse cosine of this value. (See P18 above, which is identical to this part of P79.) The answer is $\theta \$ = 42.8 degrees.

Comment: Note that the argument of the cosine has no units: when we divide the scalar product - which has units of (kiloNewtons)x(Meters), by the product of F, which has units of kiloNewtons, and then d, which has units of meters, the units divide out leaving a dimensionless ratio. It will ALWAYS be the case that the argument of a trig function is dimensionless.