Question 1

A skydiver of mass 100 kg experiences air resistance of 500 N,and an acceleration of

Accepted Answer

The answer of A skydiver of mass 100 kg experiences...

Question 2

A push on a 1-kg brick accelerates it.Neglecting friction,to equally accelerate a 10-kg brick,one would have to push with

Accepted Answer

According to Newton's Second Law of Motion (F=ma), the force required to accelerate an object is directly proportional to its mass. Therefore, to accelerate a 10-kg brick with the same acceleration as a 1-kg brick, one would need to apply 10 times the force.

Question 3

A 10-kilogram block with an initial velocity of 10 m/s slides 10 meters across a horizontal surface and comes to rest.It takes the block 2 seconds to stop.The stopping force acting on the block is about

Accepted Answer

The stopping force can be calculated using the formula F = ma, where m is mass and a is acceleration. First, find the acceleration using the formula a = (v_f - v_i) / t, where v_f is final velocity (0 m/s), v_i is initial velocity (10 m/s), and t is time (2 s). This gives a = -5 m/s². Then, F = 10 kg * (-5 m/s²) = -50 N. The magnitude of the force is 50 N.

Question 4

Suppose a particle is being accelerated through space by a 10-N force.Suddenly the particle encounters a second force of 10 N in the opposite direction from the first force.The particle with both forces acting on it

Accepted Answer

When the two forces are equal and opposite, they cancel each other out, resulting in no net force on the particle. According to Newton's first law of motion, the particle will continue moving at a constant velocity, which is the speed it had when it encountered the second force.

Question 5

The brakes of a speeding truck are slammed on and it skids to a stop.If the truck were heavily loaded so that it had twice the total mass,the skidding distance would be

Accepted Answer

The skidding distance is independent of the mass of the truck because the frictional force and the stopping distance both scale with mass, resulting in the same stopping distance regardless of mass.

Question 6

When a woman stands at rest with both feet on a scale,it reads 500 N.When she gently lifts one foot,the scale reads

Accepted Answer

The scale measures the total weight of the woman, which remains the same whether she stands on one foot or both feet.

Question 7

A 10-N block and a 1-N block lie on a horizontal frictionless table.To provide them with equal horizontal acceleration,we would have to push with

Accepted Answer

According to Newton's second law, F = ma. To achieve the same acceleration (a) for both blocks, the force (F) applied must be proportional to their masses. Since the 10-N block has 10 times the weight (and thus mass) of the 1-N block, it requires 10 times the force to achieve the same acceleration.

Question 8

A 1-kg ball is thrown at 10 m/s straight upward.Neglecting air resistance,the net force that acts on the stone when it is halfway to the top of its path is about

Accepted Answer

The net force acting on the ball at any point in its trajectory (ignoring air resistance) is due to gravity alone, which can be calculated using F = ma, where m is the mass (1 kg) and a is the acceleration due to gravity (approximately 9.8 m/s^2). This gives a force of roughly 9.8 N, which rounds to 10 N.

Question 9

An object released from rest on another planet requires one second to fall a distance of 6 meters.What is the acceleration in meters per second per second due to gravity on this planet?

Accepted Answer

The formula for acceleration due to gravity is 
a = 2d / t^2 
where a is acceleration, d is distance fallen and t is time taken. 
Substituting the given values, we get 
a = (2 x 6) / 1^2 = 12 
Therefore, the acceleration due to gravity on this planet is 12 meters per second per second. 
Hence, the correct choice is (C).

Question 10

A man weighing 800 N stands at rest on two bathroom scales so that his weight is distributed evenly over both scales.The reading on each scale is

Accepted Answer

Since the man's weight is distributed evenly on both scales, each scale will read half of his weight. Therefore, the reading on each scale will be 400 N.

Question 11

A block is dragged without acceleration in a straight-line path across a level surface by a force of 6 N.What is the force of friction between the block and the surface?

Accepted Answer

According to Newton's first law, the block will remain at rest or move with a constant velocity in a straight line unless acted upon by an unbalanced force. Since the block is not accelerating, the force of friction must be equal in magnitude and opposite in direction to the applied force of 6 N. Therefore, the force of friction is 6 N.

Question 12

A rocket becomes progressively easier to accelerate as it travels upward from the ground mainly because

Accepted Answer

As the rocket burns fuel, its mass decreases. This means that the same amount of force applied to the rocket will result in a greater acceleration because the rocket has less mass to move.

Question 13

Neglecting friction,a large block of ice and a small block of ice start sliding down an incline together.The heavier block will get to the bottom

Accepted Answer

Neglecting friction, the acceleration of the two blocks down the incline will depend only on the angle of the incline and the mass of the block, as given by the formula a = g*sin(θ), where g is the acceleration due to gravity and θ is the angle of the incline. Since both blocks start sliding together, they will have the same acceleration down the incline, regardless of their mass. Therefore, they will reach the bottom of the incline at the same time.

Question 14

A jumbo jet has a mass of 100,000 kg.The thrust for each of its four engines is 50,000 N.What is the jet's acceleration in meters per second per second when taking off?

Accepted Answer

The total thrust is 4 engines * 50,000 N = 200,000 N. Using Newton's second law, F = ma, the acceleration a = F/m = 200,000 N / 100,000 kg = 2 m/s².

Question 15

A rock is thrown vertically into the air.At the top of its path,its acceleration in meters per second per second is

Accepted Answer

At the top of its path, the rock is momentarily at rest, but the acceleration due to gravity is still acting on it, which is 9.8 m/s² downward.

Question 16

The force required to maintain an object at a constant velocity in free space is equal to

Accepted Answer

According to Newton's First Law of Motion, an object in motion will remain in motion at a constant velocity unless acted upon by an external force. Therefore, if an object is already moving at a constant velocity in free space, no external force is required to maintain that motion.

Question 17

A 1-kg rock that weighs 9.8 N is thrown straight upward at 20 m/s.Neglecting air resistance,the net force that acts on it when it is half way to the top of its path is

Accepted Answer

According to Newton's laws of motion, the net force acting on an object is given by the product of its mass and acceleration. Since the only force acting on the rock is gravity (neglecting air resistance), the net force can be calculated as:

Net force = mass x acceleration

Acceleration due to gravity = 9.8 m/s^2 (direction is downwards)

At the highest point of the trajectory, the velocity of the rock would be zero. Therefore, at the midpoint of the trajectory, the rock would be moving with a velocity of 20/2 = 10 m/s upwards. Using the equation of motion, we can calculate the height reached by the rock at this point:

v^2 = u^2 + 2as

where v = final velocity (0 m/s), u = initial velocity (10 m/s), a = acceleration (-9.8 m/s^2), and s = distance covered.

Solving for s, we get:

s = u^2/(2a) = (10^2)/(2 x (-9.8)) = 5.1 m

At this point, the net force acting on the rock would be:

Net force = mass x acceleration

= 1 kg x (-9.8 m/s^2)

= -9.8 N

Therefore, the best choice is B) 9.8 N, which is the net force acting on the rock at the midpoint of its trajectory. Note that the force is negative because it is acting in the opposite direction to the motion of the rock.

Question 18

A rock is thrown vertically into the air.At the very top of its trajectory the net force on it is

Accepted Answer

At the top of its trajectory, the rock momentarily comes to a stop before it begins to fall back down. During this brief instant, the net force acting on the rock is zero, which means that the force of gravity pulling it down is balanced by the force of the throw that sent it upward. This balance of forces corresponds to the weight of the rock, which is the force due to gravity acting on it. Therefore, the net force on the rock at the top of its trajectory is equal to its weight.

Question 19

A car traveling at 22 m/s comes to an abrupt halt in 0.1 second when it hits a tree.What is the deceleration in meters per second per second of the car?

Accepted Answer

The answer of A car traveling at 22 m/s comes...

Question 20

An object following a straight-line path at constant speed

Accepted Answer

If the object is moving at a constant speed in a straight line, then its velocity is not changing. Therefore, its acceleration is zero, which means there is no net force acting on it according to Newton's second law (F = m*a).

Quiz 4: Newtons Second Law of Motion: Force and Acceleration