PHYSM-6 A car starts to move along a straight, lev

Solution

a

To calculate the driving force when the opposing force is negligible, we use Newton's second law:

F = ma

. Here, the mass of the car is

1.1 \times 10^{3} \, \text{kg}

and the acceleration is

1.5 \, \text{m/s}^2

. Thus, the driving force is:

F = (1.1 \times 10^{3} \, \text{kg})(1.5 \, \text{m/s}^2) = 1650 \, \text{N}.

b

When the total opposing force is

600 \, \text{N}

, the net force acting on the car is given by:

F_{\text{net}} = F_{\text{driving}} - F_{\text{opposing}}.

Therefore, the driving force is:

F_{\text{driving}} = F_{\text{net}} + F_{\text{opposing}} = (1.1 \times 10^{3} \, \text{kg})(1.5 \, \text{m/s}^2) + 600 \, \text{N} = 1650 \, \text{N} + 600 \, \text{N} = 2250 \, \text{N}.

c

To calculate the distance travelled by the car in the first 10 seconds, we use the equation of motion:

s = ut + \frac{1}{2}at^2

. Since the initial speed

u = 0

, we have:

s = 0 + \frac{1}{2}(1.5 \, \text{m/s}^2)(10 \, \text{s})^2 = 0.5 \times 1.5 \times 100 = 75 \, \text{m}.

Answer

Driving force when opposing force is negligible: 1650 N; Driving force with opposing force of 600 N: 2250 N; Distance travelled in 10 s: 75 m

Key Concept

Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Equation:

F = ma

(Net force equals mass times acceleration). Kinematic equations for motion also apply.

Explanation

The calculations show how to determine the driving force under different conditions and the distance travelled using kinematic equations. This illustrates the application of Newton's laws in dynamics.

Solution

a

To determine the mean accelerating force when the golfer hits the ball, we can use Newton's second law:

F = ma

. The mass of the ball is

46 \, \text{g} = 0.046 \, \text{kg}

and the acceleration can be calculated using the change in velocity over time. The initial velocity

v_i = 0 \, \text{m/s}

and the final velocity

v_f = 70 \, \text{m/s}

, with a contact time of

0.0005 \, \text{s}

. Thus, the acceleration is:

a = \frac{v_f - v_i}{t} = \frac{70 \, \text{m/s} - 0 \, \text{m/s}}{0.0005 \, \text{s}} = 140000 \, \text{m/s}^2.

Now, substituting into the force equation:

F = (0.046 \, \text{kg})(140000 \, \text{m/s}^2) = 6440 \, \text{N}.

b

To find the mass that would exert the same force as calculated in part a, we can rearrange the force equation:

F = ma \Rightarrow m = \frac{F}{a}.

Using the force

F = 6440 \, \text{N}

and the acceleration due to gravity

g = 9.81 \, \text{m/s}^2

, we find:

m = \frac{6440 \, \text{N}}{9.81 \, \text{m/s}^2} \approx 656.5 \, \text{kg}.

1 Answer

Mean accelerating force: 6440 N; Mass exerting the same force: 656.5 kg

Key Concept

Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Equation:

F = ma

(Net force equals mass times acceleration)

Explanation

The mean accelerating force is calculated using the mass of the ball and its acceleration, while the equivalent mass is derived from the force and gravitational acceleration.

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Solution

a

To calculate the weight of the spacecraft on the Moon, we use the formula for weight:

W = mg

, where

m = 70 \, \text{kg}

and the gravitational field strength on the Moon is

g = 1.6 \, \text{N/kg}

. Thus, the weight is:

W = (70 \, \text{kg})(1.6 \, \text{N/kg}) = 112 \, \text{N}.

b

The resultant force on the spacecraft can be calculated using the formula:

F_{\text{resultant}} = F_{\text{up}} - W

, where the upwards force

F_{\text{up}} = 500 \, \text{N}

and the weight

W = 112 \, \text{N}

. Therefore, the resultant force is:

F_{\text{resultant}} = 500 \, \text{N} - 112 \, \text{N} = 388 \, \text{N}.

c

To find the acceleration of the spacecraft, we apply Newton's second law:

F = ma \Rightarrow a = \frac{F_{\text{resultant}}}{m}

. Substituting the values:

a = \frac{388 \, \text{N}}{70 \, \text{kg}} \approx 5.54 \, \text{m/s}^2.

2 Answer

Weight of spacecraft: 112 N; Resultant force: 388 N; Acceleration: 5.54 m/s²

Key Concept

Weight is the force exerted by gravity on an object, calculated as

W = mg

. The resultant force is the difference between the upward force and weight, and acceleration is derived from Newton's second law:

F = ma

.

Explanation

The weight is calculated using the mass and gravitational field strength, the resultant force is found by subtracting weight from the upward force, and acceleration is determined using the resultant force and mass.

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Solution

a

When the metal ball is dropped into the oil, it initially accelerates due to gravity. The forces acting on it are the gravitational force downward and the drag force upward, which increases with speed. As the ball accelerates, the drag force increases until it equals the weight of the ball, resulting in a net force of zero. At this point, the ball reaches terminal velocity and stops accelerating

b

To show that the metal ball reaches terminal velocity, one could measure the time it takes for the ball to fall a certain distance in the oil. If the time taken remains constant after a certain point, it indicates that the ball has reached terminal velocity, as it is no longer accelerating. This can be visually confirmed by observing the ball's motion over time

3 Answer

The ball accelerates until drag equals weight, then reaches terminal velocity; Terminal velocity can be shown by constant fall time.

Key Concept

Terminal velocity occurs when the force of gravity is balanced by the drag force, resulting in zero net force and constant velocity.

Explanation

The metal ball accelerates until the drag force equals its weight, leading to terminal velocity, which can be demonstrated by measuring constant fall time.

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Solution

a

To determine the speed of an object that travels

3 \, \mu \text{m}

in

5 \, \text{ms}

, we convert the distance to meters:

3 \, \mu \text{m} = 3 \times 10^{-6} \, \text{m}

. The speed is calculated as:

v = \frac{d}{t} = \frac{3 \times 10^{-6} \, \text{m}}{5 \times 10^{-3} \, \text{s}} = 6 \times 10^{-4} \, \text{m/s}.

b

For the distance of

6 \, \text{km}

in

3 \, \text{Ms}

, we convert

6 \, \text{km} = 6000 \, \text{m}

and

3 \, \text{Ms} = 3 \times 10^{6} \, \text{s}

. The speed is:

v = \frac{6000 \, \text{m}}{3 \times 10^{6} \, \text{s}} = 0.002 \, \text{m/s}.

c

For the distance of

8 \, \text{pm}

in

4 \, \text{ns}

, we convert

8 \, \text{pm} = 8 \times 10^{-12} \, \text{m}

and

4 \, \text{ns} = 4 \times 10^{-9} \, \text{s}

. The speed is:

v = \frac{8 \times 10^{-12} \, \text{m}}{4 \times 10^{-9} \, \text{s}} = 2 \times 10^{-3} \, \text{m/s}.

4 Answer

Speed for

3 \, \mu \text{m}

in

5 \, \text{ms}

:

6 \times 10^{-4} \, \text{m/s}

; Speed for

6 \, \text{km}

in

3 \, \text{Ms}

:

0.002 \, \text{m/s}

; Speed for

8 \, \text{pm}

in

4 \, \text{ns}

:

2 \times 10^{-3} \, \text{m/s}

Key Concept

Speed is calculated as the ratio of distance to time:

v = \frac{d}{t}

.

Explanation

The speed of an object is determined by dividing the distance traveled by the time taken, with appropriate unit conversions applied.

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Solution

a

In the context of Figure 3.18, the weight of the box acts downward on the man, while the force of the ground on the man acts upward. According to Newton's third law, the force of the box's weight is paired with the normal force exerted by the ground on the man. The body that the other force acts upon is the man

b

The direction of the other force (the normal force) is upward, opposing the weight of the box

c

The type of force involved is the normal force, which acts perpendicular to the surface of contact

5 Answer

Body: Man; Direction: Upward; Type: Normal force

Key Concept

Newton's Third Law: For every action, there is an equal and opposite reaction. Forces act in pairs.

Explanation

The forces acting on the man and the box illustrate Newton's third law, where the weight of the box and the normal force from the ground are equal and opposite.

Solution

a

To calculate the driving force when opposing forces are negligible, we use Newton's second law:

F = ma

. Here, the mass of the car is

m = 1.1 \times 10^3 \, \text{kg}

and the acceleration is

a = 1.5 \, \text{ms}^{-2}

. Thus, the driving force is:

F = (1.1 \times 10^3) \times (1.5) = 1650 \, \text{N}

b

When there is a total opposing force of 600 N, the net force acting on the car is given by:

F_{\text{net}} = F_{\text{driving}} - F_{\text{opposing}}

. Therefore, the driving force is:

F_{\text{driving}} = F_{\text{net}} + F_{\text{opposing}} = 1650 + 600 = 2250 \, \text{N}

c

To calculate the distance traveled by the car in the first 10 seconds, we use the equation of motion:

d = v_i t + \frac{1}{2} a t^2

. Since the car starts from rest,

v_i = 0

, thus:

d = 0 + \frac{1}{2} \times 1.5 \times (10)^2 = 75 \, \text{m}

Answer

Driving force (negligible opposing forces): 1650 N; Driving force (600 N opposing): 2250 N; Distance traveled: 75 m

Key Concept

Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Equation:

F = ma

(Net force equals mass times acceleration). Kinematic equations for motion under constant acceleration:

d = v_i t + \frac{1}{2} a t^2

.

Explanation

The calculations for driving force and distance traveled utilize Newton's second law and kinematic equations, demonstrating how forces and acceleration relate to motion.

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