Question 1

An object that weighs 75 N is pulled on a horizontal surface by a horizontal pull of 50 N to the right.The friction force on this object is 30 N to the left.What is the acceleration of the object? A)0.27 m/s² B)1.1 m/s² C)2.6 m/s² D)11 m/s²

Accepted Answer

The answer of An object that weighs 75 N is...

Question 2

A block is on a frictionless table,on earth.The block accelerates at 5.3 m/  When a  Horizontal force is applied to it.The block and table are set up on the Moon where the acceleration due to gravity is 1.62 m/  .A horizontal force of 5 N is now applied to the block when it is on the Moon.The acceleration imparted to the block by this force is closest to
A)2.7 m/  .
B)2.4 m/  .
C)2.9 m/  .
D)3.2 m/  .
E)3.4 m/  .

Accepted Answer

The answer of A block is on a frictionless table,on...

Question 3

A 45.0-kg person steps on a scale in an elevator.The scale reads 460 N.What is the magnitude of the acceleration of the elevator? A)4.91 m/s² B)9.81 m/s² C)46.9 m/s² D)0.206 m/s² E)0.422 m/s²

Accepted Answer

The answer of A 45.0-kg person steps on a scale...

Question 4

A certain aircraft has a mass of 300,000 kg.At a certain instant during its landing,its speed is 27.0 m/s.If the braking force is a constant 445,000 N,what is the speed of the airplane 10.0 s later?
A)10.0 m/s
B)12.2 m/s
C)14.0 m/s
D)18.0 m/s
E)20.0 m/s

Accepted Answer

To find the speed of the airplane 10.0 s later, we use the formula derived from Newton's second law, $F = ma$, where $F$ is the force applied, $m$ is the mass of the object, and $a$ is the acceleration. First, we find the acceleration using the formula $a = \frac{F}{m}$. Given that $F = -445,000$ N (negative because it's a braking force, opposing the motion) and $m = 300,000$ kg, we have:$$a = \frac{-445,000 \, \text{N}}{300,000 \, \text{kg}} = -1.483 \, \text{m/s}^2$$Next, we use the formula $v = u + at$ to find the final velocity ($v$), where $u$ is the initial velocity (27.0 m/s), $a$ is the acceleration we just calculated, and $t$ is the time (10.0 s). Plugging in the values:$$v = 27.0 \, \text{m/s} + (-1.483 \, \text{m/s}^2)(10.0 \, \text{s}) = 27.0 \, \text{m/s} - 14.83 \, \text{m/s} = 12.17 \, \text{m/s}$$Rounding to the nearest tenth gives us 12.2 m/s, which corresponds to option B.

Question 5

A 1200-kg car is pulling a 500-kg trailer along level ground.Friction of the road on the trailer is negligible.The car accelerates with an acceleration of 1.3 m/s².What is the force exerted by the car on the trailer? A)550 N B)600 N C)650 N D)700 N E)750 N

Accepted Answer

The total mass that the car is accelerating is:
m_total = m_car + m_trailer
m_total = 1200 kg + 500 kg
m_total = 1700 kg

The force exerted by the car on the trailer can be calculated using Newton's second law:
F = m_total * a
F = 1700 kg * 1.3 m/s^2
F = 2210 N

But this is the total force exerted by the car on both itself and the trailer. To find the force exerted only on the trailer, we need to subtract the force exerted on the car itself:

F_trailer = F_total - F_car
F_trailer = 2210 N - (1200 kg * 1.3 m/s^2)
F_trailer = 2210 N - 1560 N
F_trailer = 650 N

Therefore, the force exerted by the car on the trailer is 650 N. The answer is C.

Question 6

A car of mass 1100 kg that is traveling at 27 m/s starts to slow down and comes to a complete stop in 578 m.What is the magnitude of the average braking force acting on the car?
A)690 N
B)550 N
C)410 N
D)340 N

Accepted Answer

The magnitude of the average braking force can be calculated using the work-energy principle. The work done by the braking force is equal to the change in kinetic energy of the car. The initial kinetic energy (KE) of the car is $ \frac{1}{2}mv^2 $, where $ m = 1100 \, \text{kg} $ is the mass of the car and $ v = 27 \, \text{m/s} $ is its initial velocity. The final kinetic energy is 0 (since the car comes to a stop). Therefore, the work done by the braking force is equal to the initial kinetic energy:$$ \text{Work} = \Delta KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1100 \times 27^2 $$$$ \text{Work} = \frac{1}{2} \times 1100 \times 729 = 400950 \, \text{J} $$The work done is also equal to the force times the distance over which it acts (Work = Force × Distance), so:$$ 400950 = F \times 578 $$Solving for $ F $:$$ F = \frac{400950}{578} \approx 693.6 \, \text{N} $$Rounding to the nearest whole number gives a force of approximately 690 N.

Question 7

A 450-kg sports car accelerates from rest to 100 km/h in 4.80 s.What magnitude force does a 68.0 kg passenger experience during the acceleration?
A)394 N
B)82.0 N
C)342 N
D)311 N

Accepted Answer

The force experienced by the passenger can be calculated using Newton's second law, $F = ma$, where $m$ is the mass of the passenger and $a$ is the acceleration of the car. First, convert the car's acceleration from km/h to m/s: $100 \, \text{km/h} = \frac{100 \times 1000}{3600} \, \text{m/s} = 27.78 \, \text{m/s}$. The acceleration $a$ is then found by dividing the change in velocity by the time, $a = \frac{27.78 \, \text{m/s}}{4.80 \, \text{s}} = 5.79 \, \text{m/s}^2$. Applying Newton's second law for the passenger, $F = 68.0 \, \text{kg} \times 5.79 \, \text{m/s}^2 = 393.72 \, \text{N}$, which rounds to $394 \, \text{N}$.

Question 8

The figure shows an acceleration-versus-force graph for a 125-g object.What should be the value of the first tick-mark on the vertical scale,as indicated by the arrow in the figure? 
A)4
B)8
C)0.00400
D)0.00800

Accepted Answer

The answer of The figure shows an acceleration-versus-force graph for...

Question 9

The figure shows an acceleration-versus-force graph for three objects pulled by wires.If the mass of object 2 is 36 kg,what are the masses of objects 1 and 3? 
A)12 kg and 90 kg
B)72 kg and 18 kg
C)90 kg and 12 kg
D)12 kg and 72 kg

Accepted Answer

The answer of The figure shows an acceleration-versus-force graph for...

Question 10

What magnitude net force is required to accelerate a 1200-kg car uniformly from 0 m/s to 27.0 m/s in 
?
A)444 N
B)1620 N
C)3240 N
D)4360 N
E)11800 N

Accepted Answer

We can use the equation F = ma, where F is the net force, m is the mass of the car, and a is the acceleration of the car.

The acceleration of the car is:

a = (v_f - v_i) / t
a = (27.0 m/s - 0 m/s) / 11 s
a = 2.45 m/s^2

So, the net force required to accelerate the car is:

F = ma
F = (1200 kg)(2.45 m/s^2)
F = 2940 N

Therefore, the best choice is C, 3240 N, which is the closest value to 2940 N.

Question 11

The figure shows an object's acceleration-versus-force graph.What is the mass of this object? 
A)2.5 g
B)1.6 g
C)630 g
D)400,000 g

Accepted Answer

The answer of The figure shows an object's acceleration-versus-force graph.What...

Question 12

Calculate the average force a bumper would have to exert to bring a 1200-kg car to rest in 15 cm when the car had an initial speed of 2.0 m/s (about 4.5 mph). A)1.8 × 10⁴ N B)1.6 × 10⁴ N C)5.4 × 10⁴ N D)6.5 × 10⁵ N E)3.2 × 10⁴ N

Accepted Answer

The answer of Calculate the average force a bumper would...

Question 13

A block is on a frictionless table,on earth.The block accelerates at 3.0 m/ 
When a  Horizontal force is applied to it.The block and table are set up on the Moon where the acceleration due to gravity is 1.62 m/  .The weight of the block on the Moon is closest to
A)11 N.
B)9.5 N.
C)8.1 N.
D)6.8 N.
E)5.5 N.

Accepted Answer

The answer of A block is on a frictionless table,on...

Question 14

On its own,a tow truck has a maximum acceleration of 3.0 m/s².What will be its maximum acceleration when the truck is using a light horizontal chain to tow a bus of twice its own mass? A)2.5 m/s² B)2.0 m/s² C)1.5 m/s² D)1.0 m/s²

Accepted Answer

The answer of On its own,a tow truck has a...

Quiz 5: Newtons Laws of Motion

An object that weighs 75 N is pulled on a horizontal surface by a horizontal pull of 50 N to the right.The friction force on this object is 30 N to the left.What is the acceleration of the object?

A 45.0-kg person steps on a scale in an elevator.The scale reads 460 N.What is the magnitude of the acceleration of the elevator?

A certain aircraft has a mass of 300,000 kg.At a certain instant during its landing,its speed is 27.0 m/s.If the braking force is a constant 445,000 N,what is the speed of the airplane 10.0 s later?

A 1200-kg car is pulling a 500-kg trailer along level ground.Friction of the road on the trailer is negligible.The car accelerates with an acceleration of 1.3 m/s².What is the force exerted by the car on the trailer?

A car of mass 1100 kg that is traveling at 27 m/s starts to slow down and comes to a complete stop in 578 m.What is the magnitude of the average braking force acting on the car?

A 450-kg sports car accelerates from rest to 100 km/h in 4.80 s.What magnitude force does a 68.0 kg passenger experience during the acceleration?

The figure shows an acceleration-versus-force graph for a 125-g object.What should be the value of the first tick-mark on the vertical scale,as indicated by the arrow in the figure?

The figure shows an acceleration-versus-force graph for three objects pulled by wires.If the mass of object 2 is 36 kg,what are the masses of objects 1 and 3?

What magnitude net force is required to accelerate a 1200-kg car uniformly from 0 m/s to 27.0 m/s in ?

The figure shows an object's acceleration-versus-force graph.What is the mass of this object?

Calculate the average force a bumper would have to exert to bring a 1200-kg car to rest in 15 cm when the car had an initial speed of 2.0 m/s (about 4.5 mph) .

A block is on a frictionless table,on earth.The block accelerates at 3.0 m/ When a Horizontal force is applied to it.The block and table are set up on the Moon where the acceleration due to gravity is 1.62 m/ .The weight of the block on the Moon is closest to

On its own,a tow truck has a maximum acceleration of 3.0 m/s².What will be its maximum acceleration when the truck is using a light horizontal chain to tow a bus of twice its own mass?