Answers:

The free-body diagram (part a) is shown at the right.

The acceleration of the car can be computed as follows:

a = v²/R = (18.0 m/s)²/(12.0 m) = 27.0 m/s² (part b)

The net force can be found in the usual manner:

F_net = m•a = (500. kg)•(27.0 m/s²) = 13500 N (part b)

Since the center of the circle (see diagram) is above the riders, then both the net force and the acceleration vectors have an upward direction. The force of gravity is downwards, so the net force is equal to the upward force minus the downward force:

F_net = F_norm - F_grav

where F_grav = m • g = (500. kg) • (9.8 m/s/s) = 4900 N

Thus,

F_norm = F_net + F_grav = 13500 N + 4900 N = 18400 N (part c)

Answer: 1.82 m/s/s

To find acceleration, the speed and the radius must be known. The radius is given; the speed can be computed as distance per time. The dust moves a distance equivalent to 33.3 circumferences in 60 seconds. So

v = 33.3 • 2 • pi • (0.15 m) / (60 s) = 0.523 m/s.

The acceleration can now be computed using the centripetal acceleration equation:

a = v²/R = (0.523 m/s)² / (0.15 m) = 1.82 m/s/s

Answer: 69.5 N

Begin with a free-body diagram for the bucket at the bottom of its string. Since the bucket is at the bottom of the circle, the net force is upwards (inwards). Notice that the tension force must be greater than the force of gravity in order to have a net upwards (inwards) force. Also note that while F_tens is directed in the direction of the net force, F_grav is in the opposite direction. Thus, the equation can be written:

F_net = F_tens - F_grav

This equation can be rearranged to solve for tension; the expressions for F_grav (m•g) and F_net (m•v²/R) can be substituted into it:

F_tens = F_net + F_grav = m•v²/R + m•g = (2.50 kg)•(3.00 m/s)²/(0.500 m) + (2.50 kg)•(9.8 m/s²)

F_tens = 69.5 N

Answer: 254 N

At the top of the loop, both the gravitational and the normal force are directed inwards (as shown in the free-body diagram at the right). Thus, the F_net equation can be written as:

F_net = F_grav + F_norm

This equation can be rearranged and the expressions for F_grav (m•g) and F_net (m•v²/R) can be substituted into it:

F_norm = F_net - F_grav = m • v² / R - m • g

F_norm = (53.5 kg) • (10.3 m/s)² / (7.29 m) - (53.5 kg) • (9.8 m/s²)

F_norm = 254 N

Answer: 3.40

As usual, begin with a free-body diagram. The gravity force is balanced by (and equal to) the normal force and the force of friction is the net force. The solution then begins by equating m • a to F_frict and carrying out the customary substitutions and algebra steps (using the fact that a = v² / R and F_frict = µ • F_norm and F_grav = m • g).

The mass now cancels and the equation can be rearranged to solve for mu:

(This is not a likely mu value and as such the car does not have much of a chance negotiating the turn at this speed.)

Answer: 42.8 mi/hr (19.1 m/s)

The feeling of weightlessness is experienced when a person is under the sole influence of gravity. So the free-body diagram for such a person would reveal only one force - gravity, pointing straight down. In such a case, the seat does not push upward upon the person. The car has passed over the crest of the hill at such a high speed that it is momentarily raised off the roadway and is unable to support the weight of its passengers. The force of gravity is the net force and the following equation can be written:

The mass can be canceled from both sides of the equation (sigh!) and the expression for acceleration can be substituted into the equation, yielding:

Rearrangement and substitution yields the solution:

This value for speed can be converted to mi/hr using the equivalency: 2.24 mi/hr = 1m/s; the car is traveling with a speed of 42.8 mi/hr.

Answer: 1.1 x 10³ N

The scale reading is a measure of the amount of normal force acting upon the person. As is the usual case, begin with a free-body diagram (shown at the right). Since the acceleration is upwards, the upwards normal force (F_norm) is greater than the downwards gravitational force (F_grav). From the diagram, the F_net equation can be written:

F_net = F_norm - F_grav

Algebraic rearrangement of this equation, combined with the understanding that F_net = m•a and F_grav = m•g yields the following solution:

F_norm = F_net + F_grav = m•a + m•g = (72 kg) • (5.4 m/s²) + (72 kg) • (9.8 m/s²)

F_norm = 1.1 x 10³ N

Answer: 3.96 x 10³ N

As usual, begin with a free-body diagram (shown at the right). From the diagram, the equation relating net force to the individual forces can be written:

F_net = F_norm + F_grav

Since F_net = m•a = m•v²/R and F_grav = m•g, substitutions can be made and the equation can be rewritten:

m•v²/R = F_norm + m•g

Rearranging to solve for F_norm yields:

F_norm = m•v²/R - m•g = (62.6 kg) • (64.1 m/s)² / (56.2 m) - (62.6 kg) • (9.8 m/s²)

F_norm = 3.96 x 10³ N

Answer: 0.61 yr

This question explores the R-T relationship. As such, Kepler's third law - that the T²/R³ ratio is the same for all the planets - must be utilized.

Rearranging this equation to solve for the period of the planet yields the equation:

T_Venus = SQRT(0.373 year² ) = 0.61 years

Answer: 3.6•10^-47 N

Use Newton's universal gravitation equation:

F_grav = G • m₁ • m₂ /R²

F_grav = (6.67•10^-11 N•m²/kg²) • (1.67 x 10^-27 kg) • (9.11x10^-31 kg) /(5.3 x 10^-11 m)²

F_grav = 3.6•10^-47 N

Answers:

The first step is to find the separation distance from the center of the earth - this is R in most of the equations. The altitude of the satellite is 15000 mi or 2.41•10⁷ m; this distance must be added to the radius of the earth to determine the separation distance. The radius of the earth is 6.38•10⁶ m; the separation distance is 3.05•10⁷ m. The orbital velocity can be computed using the orbital velocity equation:

v = SQRT (1.31•10⁷ m²/s²) = 3615 m/s

The orbital period can now be calculated using the v = 2•pi•R/T equation or the Kepler's second law equation [ T²/R³ = 4•pi²/(GM_earth)]. Starting with the first equation, algebra yields:

T = (2•pi•R)/v = [2 • pi • (3.05•10⁷ m) / (3615 m/s) ] = 5.30 • 10⁴ s = 14.7 hr