Question 1

A 10-kg mass and a 6.0-kg mass are connected by a light string over a massless,frictionless pulley.If g = 9.8 m/s²,what is the acceleration of the system when released?

Accepted Answer

The acceleration of the system can be found using the formula: a = (m1 - m2) * g / (m1 + m2), where m1 = 10 kg and m2 = 6 kg. Substituting the values, a = (10 - 6) * 9.8 / (10 + 6) = 4 * 9.8 / 16 = 2.45 m/s², which rounds to 2.5 m/s².

Question 2

Plane A is flying at 600 mph in the northeast direction relative to the Earth.Plane B is flying at 500 mph in the north direction relative to the Earth.What is the speed of Plane B as observed from Plane A?

Accepted Answer

To find the relative speed of Plane B as observed from Plane A, we need to subtract the velocity of Plane A (in the northeast direction) from the velocity of Plane B (in the north direction). 
Let's break the velocity vector of Plane A into its northern and eastern components:
Velocity of Plane A = 600 mph (northeast)
= 600/√2 mph (north) + 600/√2 mph (east) (using trigonometry)

Therefore, the velocity of Plane B as observed from Plane A is:
Velocity of Plane B (relative to A) = Velocity of Plane B - Velocity of Plane A
= 500 mph (north) - (600/√2) mph (north)
= (500 - 600/√2) mph (north)
= 431 mph (north) (approx)

Hence, the answer is (B) 431 mph.

Question 3

A plane is moving due north,directly towards its destination.Its airspeed is 200 mph.A constant breeze is blowing from west to east at 40 mph.How long will it take for the plane to travel 200 miles north?

Accepted Answer

The plane is not traveling in a straight line due to the effect of the wind. The wind will push the plane slightly towards the east as it travels north, causing it to travel a longer distance than just the 200 miles north. The exact time it will take depends on the exact angle of the wind and the speed of the plane, so we cannot determine the answer with the information given.

Question 4

A 2000-kg sailboat experiences an eastward force of 2000 N by the ocean tide and a wind force against its sails with magnitude of 6000 N directed toward the northwest (45° N of W).What is the direction of the resultant acceleration?

Accepted Answer

The answer of A 2000-kg sailboat experiences an eastward force...

Question 5

A boat travels upstream and after one hour has gone 10 km.The boat next travels downstream and after one hour has gone 16 km.If the boat's speed relative to the water is constant,what is the speed of the current in the river?

Accepted Answer

The answer of A boat travels upstream and after one...

Question 6

A 12.0-kg mass is placed on a 25.0° incline and friction keeps it from sliding.The coefficient of static friction in this case is 0.580,and the coefficient of sliding friction is 0.520.The mass is given a shove causing it to slide down the incline.What is the frictional force while the mass is sliding?

Accepted Answer

The answer of A 12.0-kg mass is placed on a...

Question 7

A river flows due east at 2.0 m/s.A boat crosses the 300 m-wide river by maintaining a constant velocity of 10 m/s due north relative to the water.If no correction is made for the current,how far downstream does the boat move by the time it reaches the far shore?

Accepted Answer

The boat's velocity can be broken down into two components: a northward velocity of 10 m/s and an eastward velocity due to the river's current of 2 m/s. 
The time it takes for the boat to cross the river is given by t=300m/10m/s=30s. 
During this time, the boat will also drift downstream due to the river's current. The distance it drifts downstream is given by d=vt=(2m/s)(30s)=60m. 
Therefore, the answer is C, 60m.

Question 8

Two ropes are attached to a 40-kg object.The first rope applies a force of 50 N and the second,40 N.If the two ropes are perpendicular to each other,what is the resultant magnitude of the acceleration of the object?

Accepted Answer

To find the resultant acceleration, we need to first find the net force acting on the object.
We can use Pythagoras theorem to find the magnitude of the net force:
$F_{net} = \sqrt{(50	ext{ N})^2 + (40	ext{ N})^2} = 64	ext{ N}$
Using Newton's second law, we can find the acceleration:
$F_{net} = ma$
$64	ext{ N} = 40	ext{ kg} 	imes a$
$a = 1.6	ext{ m/s}^2$
Therefore, the resultant magnitude of the acceleration is 1.6 m/s^2, which is answer choice B.

Question 9

A 2000-kg sailboat experiences an eastward force of 2000 N by the ocean tide and a wind force against its sails with magnitude of 6000 N directed toward the northwest (45° N of W).What is the magnitude of the resultant acceleration?

Accepted Answer

First, we need to resolve the wind force vector into east and north components using trigonometry: 
North Force = 6000 N * sin(45°) = 4243 N 
East Force = 6000 N * cos(45°) = 4243 N 
Next, we can calculate the net force in the eastward direction: 
Net eastward force = 4243 N - 2000 N = 2243 N 
Then, we can calculate the net force in the northward direction: 
Net northward force = 4243 N 
Since there is no force acting in the upward or downward direction, we can use the Pythagorean theorem to find the magnitude of the resultant acceleration: 
Resultant acceleration = √[(2243 N)^2 + (4243 N)^2] / (2000 kg) = 2.4 m/s^2

Question 10

A 15-kg block rests on a level,frictionless surface and is attached by a light string to a 10-kg hanging mass where the string passes over a massless frictionless pulley.If g = 9.8 m/s²,what is the tension in the connecting string?

Accepted Answer

Since the surface is frictionless, there is no horizontal or vertical component of the tension force, thus the tension force is simply equal to the weight of the hanging mass: 
Tension = (mass of hanging mass) * g = 10 kg * 9.8 m/s^2 = 98 N
Since the pulley is massless and frictionless, the tension force on both sides of the pulley must be equal, thus the tension force on the block is also 98 N.

Question 11

A 150-N sled is pulled up a 28° slope at a constant speed by a force of 120 N.What is the coefficient of kinetic friction between sled and slope?

Accepted Answer

The force acting parallel (up the slope) is 120 N and the perpendicular force is mg (where g is the acceleration due to gravity and m is the mass). The perpendicular force is m*g*cos(28°) and therefore, the mass is (m*g*cos(28°))/g = m*cos(28°). 
Since the sled is moving at a constant speed, friction must be equal to the parallel force (120 N), and the frictional force is given as f = μ*mg*cos(28°). 
Equating f to 120 N, we get μ = 120/(m*g*cos(28°)). Substituting the value of m*cos(28°) in terms of m, we get μ = 120/(m*g*cos(28°)) = 120/(m*g*cos(28°)) = 120/(150*cos(28°)) = 0.374. 
Therefore, the coefficient of kinetic friction between sled and slope is 0.37 (option C).

Question 12

A boxcar of mass 200 tons at rest becomes uncoupled on a 1.5° grade.If the track is considered to be frictionless,what speed does the boxcar have after 10 seconds?

Accepted Answer

We can use the equation: $d = \frac{1}{2}at^2$ where $d$ is the distance travelled, $a$ is the acceleration and $t$ is time.

First, we need to find the acceleration:
$$\sin(1.5^\circ) = \frac{h}{l}$$
where $h$ is the height gain and $l$ is the horizontal distance travelled. Therefore, for a 200 ton boxcar, the height gain is:
$$h = \sin(1.5^\circ) 	imes 200	ext{ tons} 	imes 1000	ext{ kg/ton} 	imes 9.81	ext{ m/s}^2 = 534.9	ext{ m}$$
The horizontal distance travelled is:
$$l = \cos(1.5^\circ) 	imes d$$
where $d$ is the total distance travelled. We know that $d = \frac{1}{2}at^2$, so we need to solve for $d$. 
The final velocity, $v$, can also be expressed as $v = at$, so we can eliminate $a$ between the two equations to get:
$$v = \sqrt{2gd	an(1.5^\circ)} = \sqrt{2 	imes 9.81	ext{ m/s}^2 	imes 534.9	ext{ m} 	imes 	an(1.5^\circ)} = 9.81	ext{ m/s} 	imes 11.13 = 109.29	ext{ m/s}$$
Therefore, the horizontal distance travelled is:
$$l = \cos(1.5^\circ) 	imes \frac{1}{2} 	imes 9.81	ext{ m/s}^2 	imes (10	ext{ s})^2 = 480.74	ext{ m}$$
So the total distance travelled is:
$$d = 2l = 961.48	ext{ m}$$
Finally, we can calculate the velocity:
$$v = \sqrt{2gd	an(1.5^\circ)} = \sqrt{2 	imes 9.81	ext{ m/s}^2 	imes 534.9	ext{ m} 	imes 	an(1.5^\circ)} \approx 2.6	ext{ m/s}$$
Answer: C.

Question 13

A plane is moving due north,directly towards its destination.Its airspeed is 200 mph.A constant breeze is blowing from west to east at 42 mph.In which direction is the plane pointed?

Accepted Answer

To solve this problem, we need to use vector addition. Let's break down the velocities into components:

- Airspeed (200 mph) directly North: 0 mph West, 200 mph North
- Breeze (42 mph) West to East: 42 mph East, 0 mph North

We can add these two vectors to get the resulting velocity:

- Resulting velocity: 42 mph East, 200 mph North

To find the direction of this velocity, we can use trigonometry. The angle between this velocity and due North is:
theta = arctan(42/200) = 12.1°

Therefore, the plane is pointed 12° West of North, which corresponds to choice B.

Question 14

A 12.0-kg mass is placed on a 25.0° incline and friction keeps it from sliding.The coefficient of static friction in this case is 0.580 and the coefficient of sliding friction is 0.520.The mass is given a shove causing it to slide down the incline.Taking down the incline as positive,what is the acceleration of the mass while it is sliding?

Accepted Answer

The answer of A 12.0-kg mass is placed on a...

Question 15

As I pull a rope connected to a box at constant speed up a frictionless slope,pulling parallel to the slope,the tension in the rope will be:

Accepted Answer

When pulling a box up a frictionless slope at constant speed, the tension in the rope must balance the component of the box's weight parallel to the slope. Since this component is less than the total weight of the box, the tension will be less than the weight of the box.

Question 16

A plane is moving due north,directly towards its destination.Its airspeed is 200 mph.A constant breeze is blowing from west to east at 60 mph.At what rate is the plane moving north?

Accepted Answer

The answer of A plane is moving due north,directly towards...

Question 17

The accelerating force of the wind on a small 250-kg sailboat is 707 N northeast.If the drag of the keel is 500 N acting west,what is the acceleration of the boat?

Accepted Answer

The answer of The accelerating force of the wind on...

Question 18

Plane A is flying at 600 mph in the northeast direction relative to the Earth.Plane B is flying at 500 mph in the north direction relative to the Earth.What is the direction of motion of Plane B as observed from Plane A?

Accepted Answer

The answer of Plane A is flying at 600 mph...

Question 19

A 12-kg mass is placed on a 25.0° incline,and friction keeps it from sliding.The coefficient of static friction in this case is 0.580,and the coefficient of sliding friction is 0.520.What is the frictional force in this situation?

Accepted Answer

Since the mass is not sliding, the frictional force is equal and opposite to the component of the weight that is parallel to the incline. The weight parallel to the incline can be found using trigonometry:

weight parallel = mg sin(25.0°) = (12 kg)(9.81 m/s^2)(sin 25.0°) = 51.0 N

The static frictional force will be equal and opposite to this, up to a maximum value of μsN, where N is the normal force. The normal force is equal to the weight component perpendicular to the incline:

N = mg cos(25.0°) = (12 kg)(9.81 m/s^2)(cos 25.0°) = 109.9 N

Thus, the maximum possible static frictional force is:

μsN = (0.580)(109.9 N) = 63.7 N

Since the weight parallel is less than this maximum, the actual static frictional force will be equal and opposite to the weight parallel:

frictional force = weight parallel = 51.0 N

Note that the coefficient of sliding friction is not relevant here since the mass is not sliding.

Question 20

A 250-kg crate is placed on an adjustable inclined plane.If the crate slides down the incline with an acceleration of 0.25 m/s² when the incline angle is 25°,then what should the incline angle be for the crate to slide down the plane at constant speed? Note: g = 9.8 m/s².

Accepted Answer

For the crate to slide at constant speed, the net force must be zero, meaning the component of gravitational force along the incline equals the frictional force. The angle where this occurs is slightly less than the angle causing acceleration. Calculating using the formula for acceleration on an incline, we find the angle to be approximately 24°.

Quiz 4: The Laws of Motion: Part A

A 10-kg mass and a 6.0-kg mass are connected by a light string over a massless,frictionless pulley.If g = 9.8 m/s²,what is the acceleration of the system when released?

Plane A is flying at 600 mph in the northeast direction relative to the Earth.Plane B is flying at 500 mph in the north direction relative to the Earth.What is the speed of Plane B as observed from Plane A?

A plane is moving due north,directly towards its destination.Its airspeed is 200 mph.A constant breeze is blowing from west to east at 40 mph.How long will it take for the plane to travel 200 miles north?

A 2000-kg sailboat experiences an eastward force of 2000 N by the ocean tide and a wind force against its sails with magnitude of 6000 N directed toward the northwest (45° N of W) .What is the direction of the resultant acceleration?

A boat travels upstream and after one hour has gone 10 km.The boat next travels downstream and after one hour has gone 16 km.If the boat's speed relative to the water is constant,what is the speed of the current in the river?

A river flows due east at 2.0 m/s.A boat crosses the 300 m-wide river by maintaining a constant velocity of 10 m/s due north relative to the water.If no correction is made for the current,how far downstream does the boat move by the time it reaches the far shore?

Two ropes are attached to a 40-kg object.The first rope applies a force of 50 N and the second,40 N.If the two ropes are perpendicular to each other,what is the resultant magnitude of the acceleration of the object?

A 2000-kg sailboat experiences an eastward force of 2000 N by the ocean tide and a wind force against its sails with magnitude of 6000 N directed toward the northwest (45° N of W) .What is the magnitude of the resultant acceleration?

A 15-kg block rests on a level,frictionless surface and is attached by a light string to a 10-kg hanging mass where the string passes over a massless frictionless pulley.If g = 9.8 m/s²,what is the tension in the connecting string?

A 150-N sled is pulled up a 28° slope at a constant speed by a force of 120 N.What is the coefficient of kinetic friction between sled and slope?

A boxcar of mass 200 tons at rest becomes uncoupled on a 1.5° grade.If the track is considered to be frictionless,what speed does the boxcar have after 10 seconds?

A plane is moving due north,directly towards its destination.Its airspeed is 200 mph.A constant breeze is blowing from west to east at 42 mph.In which direction is the plane pointed?

As I pull a rope connected to a box at constant speed up a frictionless slope,pulling parallel to the slope,the tension in the rope will be:

A plane is moving due north,directly towards its destination.Its airspeed is 200 mph.A constant breeze is blowing from west to east at 60 mph.At what rate is the plane moving north?

The accelerating force of the wind on a small 250-kg sailboat is 707 N northeast.If the drag of the keel is 500 N acting west,what is the acceleration of the boat?

Plane A is flying at 600 mph in the northeast direction relative to the Earth.Plane B is flying at 500 mph in the north direction relative to the Earth.What is the direction of motion of Plane B as observed from Plane A?

A 12-kg mass is placed on a 25.0° incline,and friction keeps it from sliding.The coefficient of static friction in this case is 0.580,and the coefficient of sliding friction is 0.520.What is the frictional force in this situation?

A 250-kg crate is placed on an adjustable inclined plane.If the crate slides down the incline with an acceleration of 0.25 m/s² when the incline angle is 25°,then what should the incline angle be for the crate to slide down the plane at constant speed? Note: g = 9.8 m/s².

Quiz 4: The Laws of Motion: Part A

A 10-kg mass and a 6.0-kg mass are connected by a light string over a massless,frictionless pulley.If g = 9.8 m/s2,what is the acceleration of the system when released?

Plane A is flying at 600 mph in the northeast direction relative to the Earth.Plane B is flying at 500 mph in the north direction relative to the Earth.What is the speed of Plane B as observed from Plane A?

A plane is moving due north,directly towards its destination.Its airspeed is 200 mph.A constant breeze is blowing from west to east at 40 mph.How long will it take for the plane to travel 200 miles north?

A 2000-kg sailboat experiences an eastward force of 2000 N by the ocean tide and a wind force against its sails with magnitude of 6000 N directed toward the northwest (45° N of W) .What is the direction of the resultant acceleration?

A boat travels upstream and after one hour has gone 10 km.The boat next travels downstream and after one hour has gone 16 km.If the boat's speed relative to the water is constant,what is the speed of the current in the river?

A river flows due east at 2.0 m/s.A boat crosses the 300 m-wide river by maintaining a constant velocity of 10 m/s due north relative to the water.If no correction is made for the current,how far downstream does the boat move by the time it reaches the far shore?

Two ropes are attached to a 40-kg object.The first rope applies a force of 50 N and the second,40 N.If the two ropes are perpendicular to each other,what is the resultant magnitude of the acceleration of the object?

A 2000-kg sailboat experiences an eastward force of 2000 N by the ocean tide and a wind force against its sails with magnitude of 6000 N directed toward the northwest (45° N of W) .What is the magnitude of the resultant acceleration?

A 15-kg block rests on a level,frictionless surface and is attached by a light string to a 10-kg hanging mass where the string passes over a massless frictionless pulley.If g = 9.8 m/s2,what is the tension in the connecting string?

A 150-N sled is pulled up a 28° slope at a constant speed by a force of 120 N.What is the coefficient of kinetic friction between sled and slope?

A boxcar of mass 200 tons at rest becomes uncoupled on a 1.5° grade.If the track is considered to be frictionless,what speed does the boxcar have after 10 seconds?

A plane is moving due north,directly towards its destination.Its airspeed is 200 mph.A constant breeze is blowing from west to east at 42 mph.In which direction is the plane pointed?

As I pull a rope connected to a box at constant speed up a frictionless slope,pulling parallel to the slope,the tension in the rope will be:

A plane is moving due north,directly towards its destination.Its airspeed is 200 mph.A constant breeze is blowing from west to east at 60 mph.At what rate is the plane moving north?

The accelerating force of the wind on a small 250-kg sailboat is 707 N northeast.If the drag of the keel is 500 N acting west,what is the acceleration of the boat?

Plane A is flying at 600 mph in the northeast direction relative to the Earth.Plane B is flying at 500 mph in the north direction relative to the Earth.What is the direction of motion of Plane B as observed from Plane A?

A 12-kg mass is placed on a 25.0° incline,and friction keeps it from sliding.The coefficient of static friction in this case is 0.580,and the coefficient of sliding friction is 0.520.What is the frictional force in this situation?

A 250-kg crate is placed on an adjustable inclined plane.If the crate slides down the incline with an acceleration of 0.25 m/s2 when the incline angle is 25°,then what should the incline angle be for the crate to slide down the plane at constant speed? Note: g = 9.8 m/s2.

A 10-kg mass and a 6.0-kg mass are connected by a light string over a massless,frictionless pulley.If g = 9.8 m/s²,what is the acceleration of the system when released?

A 15-kg block rests on a level,frictionless surface and is attached by a light string to a 10-kg hanging mass where the string passes over a massless frictionless pulley.If g = 9.8 m/s²,what is the tension in the connecting string?

A 250-kg crate is placed on an adjustable inclined plane.If the crate slides down the incline with an acceleration of 0.25 m/s² when the incline angle is 25°,then what should the incline angle be for the crate to slide down the plane at constant speed? Note: g = 9.8 m/s².