Question 1

If F = 4.0 N and m = 2.0 kg,what is the magnitude a of the acceleration for the block shown below? The surface is frictionless. A)5.3 m/s² B)4.4 m/s² C)3.5 m/s² D)6.2 m/s² E)8.4 m/s²

Accepted Answer

Newton's second law states that F = ma, where F is the net force acting on an object, m is the object's mass, and a is the acceleration. In this case, the net force is given as F = 4 N and the mass as m = 2 kg. Thus, we can find the magnitude of the acceleration by rearranging the equation to a = F/m:

a = F/m = 4 N / 2 kg = 2 m/s^2

Therefore, the magnitude of the acceleration is 2 m/s^2, which is closest to choice C (3.5 m/s^2).

Question 2

If α = 40°,β = 60°,and M = 4.0 kg,determine the tension in string 1. 
A)15 N
B)22 N
C)17 N
D)20 N
E)36 N

Accepted Answer

D

Question 3

If M = 6.0 kg,what is the tension in string 1? 
A)39 N
B)34 N
C)29 N
D)44 N
E)51 N

Accepted Answer

E

Question 4

The only two forces acting on a body have magnitudes of 20 N and 35 N and directions that differ by 80°.The resulting acceleration has a magnitude of 20 m/s².What is the mass of the body? A)2.4 kg B)2.2 kg C)2.7 kg D)3.1 kg E)1.5 kg

Accepted Answer

First, find the x and y components of each force by using trigonometry:
- 20 N force: x-component = 20 cos 80° ≈ 3.92 N, y-component = 20 sin 80° ≈ 19.37 N 
- 35 N force: x-component = 35 cos 260° ≈ -30.31 N, y-component = 35 sin 260° ≈ -15.99 N (260° is the same as -80°)
Next, find the total x and y components of the forces by adding the x and y components separately: 
- Total x-component = 3.92 N - 30.31 N ≈ -26.39 N
- Total y-component = 19.37 N - 15.99 N ≈ 3.38 N
Use the Pythagorean theorem to find the magnitude of the total force: 
- Total force = sqrt((-26.39 N)^2 + (3.38 N)^2) ≈ 26.64 N
Now that we have the total force, we can use F=ma to find the mass: 
- F = 26.64 N
- a = 20 m/s^2 (given)
- rearranging the formula gives m = F/a = 26.64 N / 20 m/s^2 ≈ 1.32 kg
Therefore, the closest answer choice is B) 2.2 kg.

Question 5

In the figure,if the tension in string 1 is 34 N and the tension in string 2 is 24 N,what is the mass of the object shown?
​ 
​
A)7.3 kg
B)5.5 kg
C)1.8 kg
D)3.7 kg
E)4.5 kg

Accepted Answer

The answer of In the figure,if the tension in string...

Question 6

Two forces are the only forces acting on a 3.0-kg object which moves with an acceleration of 3.0 m/s² in the positive y direction.If one of the forces acts in the positive x direction and has a magnitude of 8.0 N,what is the magnitude of the other force? A)12 N B)14 N C)16 N D)18 N E)22 N

Accepted Answer

The answer of Two forces are the only forces acting...

Question 7

If α = 40° and the tension in string 2 is 30 N,determine M. 
A)3.4 kg
B)3.6 kg
C)2.6 kg
D)4.9 kg
E)7.5 kg

Accepted Answer

Using the equilibrium equation for forces in the vertical direction:
T1sinα - Mgsinθ - T2sinα = 0
T1sinα = T2sinα + Mgsinθ
T1 = T2 + Mgtanθ
M = (T1-T2)/g tan θ
M = (T1-T2)/(9.81 tan 40°)
M = (T1 - 30)/(9.81 x 0.839)
M = (T1 - 30)/8.21

Given T1 = 50 N
M = (50 - 30)/8.21
M = 2.43 kg

Therefore, the closest option is C) 2.6 kg.

Question 8

A 5.0-kg mass is attached to the ceiling of an elevator by a rope whose mass is negligible.What force does the mass exert on the rope when the elevator has an acceleration of 4.0 m/s² upward? A)69 N downward B)29 N downward C)49 N downward D)20 N downward E)19 N downward

Accepted Answer

The answer of A 5.0-kg mass is attached to the...

Question 9

The tension in a string from which a 4.0-kg object is suspended in an elevator is equal to 44 N.What is the acceleration of the elevator? A)11 m/s² upward B)1.2 m/s² upward C)1.2 m/s² downward D)10 m/s² upward E)2.4 m/s² downward

Accepted Answer

The tension in the string (T) is given by the equation T = mg + ma, where m is the mass of the object, g is the acceleration due to gravity (9.8 m/s^2), and a is the acceleration of the elevator. Rearranging for a gives a = (T/m) - g. Substituting the given values, a = (44N / 4.0kg) - 9.8 m/s^2 = 11 m/s^2 - 9.8 m/s^2 = 1.2 m/s^2 upward.

Question 10

An object of unknown weight is suspended as shown.The tension in rope 1 is 25 lb,and the tension in rope 2 is 31 lb.What is the weight of the suspended object? 
A)36 lb
B)33 lb
C)41 lb
D)39 lb
E)56 lb

Accepted Answer

The weight of the object is equal to the sum of the tensions in the two ropes. Therefore, weight = tension in rope 1 + tension in rope 2 = 25 lb + 31 lb = 56 lb. So, the correct option is D.

Question 11

A 3.0-kg block slides on a frictionless 20° inclined plane.A force of 16 N acting parallel to the incline and up the incline is applied to the block.What is the acceleration of the block? A)2.0 m/s² down the incline B)5.3 m/s² up the incline C)2.0 m/s² up the incline D)3.9 m/s² down the incline E)3.9 m/s² up the incline

Accepted Answer

The answer of A 3.0-kg block slides on a frictionless...

Question 12

At an instant when a 4.0-kg object has an acceleration equal to M/s²,one of the two forces acting on the object is known to be n.Determine the magnitude of the other force acting on the object. A)2.0 N B)13 N C)18 N D)1.7 N E)20 N

Accepted Answer

Using Newton's second law, F = ma, we can solve for the unknown force:

F = ma - known force
F = 4.0 kg x 11e7ff9_2a7b_9e19_a768_39118c1701d2_TB2295_11 M/s^2 - 11ea7ff9_2a7b_9e1a_a768_0feaa184164f_TB2295_11 N
F = 44 N - 13 N
F = 31 N

Therefore, the magnitude of the other force acting on the object is 31 N, which corresponds to choice B.

Question 13

A block is pushed up a frictionless 30° incline by an applied force as shown.If F = 25 N and M = 3.0 kg,what is the magnitude of the resulting acceleration of the block? A)2.3 m/s² B)4.6 m/s² C)3.5 m/s² D)2.9 m/s² E)5.1 m/s²

Accepted Answer

The answer of A block is pushed up a frictionless...

Question 14

A 5.0-kg mass is suspended by a string from the ceiling of an elevator that is moving upward with a speed which is decreasing at a constant rate of 2.0 m/s in each second.What is the tension in the string supporting the mass?
A)49 N
B)39 N
C)59 N
D)10 N
E)42 N

Accepted Answer

The force diagram of the mass includes the force of gravity acting downward and the tension force acting upward. Since the elevator is accelerating downwards, the net force acting on the mass is in the downward direction. Thus, the tension force must be less than the force of gravity. 
Using Newton's second law, we know that F = ma. The mass is not accelerating, so the net force must be 0. Therefore, the tension force minus the force of gravity must be equal to 0. 
T - mg = 0
T = mg 
T = (5.0 kg) * (9.8 m/s^2) 
T = 49 N 
However, since the elevator is accelerating downwards, the tension force will be less than 49 N. The elevator's acceleration is 2.0 m/s^2, so we can subtract that from the acceleration due to gravity: 
a_net = g - a_elevator 
a_net = (9.8 m/s^2) - (2.0 m/s^2) 
a_net = 7.8 m/s^2 
Now we can find the tension force: 
T = ma_net 
T = (5.0 kg) * (7.8 m/s^2) 
T = 39 N 
Therefore, the tension in the string supporting the mass is 39 N.

Question 15

If M = 1.1 kg,what is the tension in string 1? 
A)54 N
B)47 N
C)40 N
D)62 N
E)57 N

Accepted Answer

The answer of If M = 1.1 kg,what is the...

Question 16

The horizontal surface on which the block slides is frictionless.If F = 20 N and M = 5.0 kg,what is the magnitude of the resulting acceleration of the block? A)5.3 m/s² B)6.2 m/s² C)7.5 m/s² D)4.7 m/s² E)3.2 m/s²

Accepted Answer

The net force on the block is given by:
Fnet = F - f
Since the surface is frictionless, f = 0
Therefore, Fnet = F = 20 N
Using Newton's second law,
Fnet = ma
a = Fnet/m
Substituting the given values, we get:
a = 20/5 = 4 m/s^2
Therefore, the magnitude of acceleration of the block is 4 m/s^2, which is closest to option C, 7.5 m/s^2. The other options are too far off the calculated value.

Question 17

If the only forces acting on a 2.0-kg mass are N and N,what is the magnitude of the acceleration of the particle? A)1.5 m/s² B)6.5 m/s² C)4.7 m/s² D)9.4 m/s² E)7.2 m/s²

Accepted Answer

The net force acting on the particle is the vector sum of the given forces: 
net force = 11ea7ff9_2a7b_28e7_a768_27981424a318_TB2295_11N + 11ea7ff9_2a7b_28e8_a768_630a229931de_TB2295_11N = 22N
Using Newton's second law, F=ma, we can solve for the acceleration: 
a = F/m = 22N / 2.0kg = 11m/s^2
Therefore, the magnitude of the acceleration is 4.7 m/s^2 (option C).

Question 18

A person weighing 0.70 kN rides in an elevator that has an upward acceleration of 1.5 m/s².What is the magnitude of the force of the elevator floor on the person? A)0.11 kN B)0.81 kN C)0.70 kN D)0.59 kN E)0.64 kN

Accepted Answer

We can use Newton's second law, which states that the net force on an object is equal to the product of its mass and acceleration:

F_net = m*a

In this case, the person has a mass of:

m = 0.70 kN / 9.81 m/s^2 = 0.0713 kg

The upward acceleration of the elevator is:

a = 1.5 m/s^2

Therefore, the net force on the person is:

F_net = m*a = (0.0713 kg)*(1.5 m/s^2) = 0.107 kN

Since the person is in contact with the floor of the elevator, the force exerted by the floor on the person must be equal and opposite to the net force, so:

F_floor = -F_net = -0.107 kN

However, we want the magnitude of the force, so we take the absolute value:

|F_floor| = 0.107 kN ≈ 0.81 kN (to two significant figures)

Question 19

If M = 2.0 kg,what is the tension in string 1? 
A)1.2 N
B)11 N
C)34 N
D)3.5 N
E)40 N

Accepted Answer

The tension in string 1 can be found by calculating the net force acting on the system in the vertical direction. Considering upward direction as positive,
T1 - M*g - T2*sin(30) = M*a
where a is the acceleration of the system, g is acceleration due to gravity, T2 is tension in the second string and sin(30) is due to the angle between string 2 and the horizontal. As the system is in equilibrium, a = 0. Therefore,
T1 = M*g + T2*sin(30)
Now, T2 can be found from the equation of torque about the hinge point,
T2*cos(30) * L = M*g * L/2
where L is the length of the horizontal bar. Solving this equation for T2,
T2 = M*g/2cos(30)
Substituting this value in the above equation for T1,
T1 = M*g + M*g/4 = 5M*g/4 = 5*2*9.81/4 = 34.26 N (approx). Therefore, the best choice is C (34 N).

Question 20

A 5.0-kg object is suspended by a string from the ceiling of an elevator that is accelerating downward at a rate of 2.6 m/s².What is the tension in the string? A)49 N B)36 N C)62 N D)13 N E)52 N

Accepted Answer

The answer of A 5.0-kg object is suspended by a...

Quiz 5: The Laws of Motion

If F = 4.0 N and m = 2.0 kg,what is the magnitude a of the acceleration for the block shown below? The surface is frictionless.

If α = 40°,β = 60°,and M = 4.0 kg,determine the tension in string 1.

If M = 6.0 kg,what is the tension in string 1?

The only two forces acting on a body have magnitudes of 20 N and 35 N and directions that differ by 80°.The resulting acceleration has a magnitude of 20 m/s2.What is the mass of the body?

In the figure,if the tension in string 1 is 34 N and the tension in string 2 is 24 N,what is the mass of the object shown? ​ ​

Two forces are the only forces acting on a 3.0-kg object which moves with an acceleration of 3.0 m/s2 in the positive y direction.If one of the forces acts in the positive x direction and has a magnitude of 8.0 N,what is the magnitude of the other force?

If α = 40° and the tension in string 2 is 30 N,determine M.

A 5.0-kg mass is attached to the ceiling of an elevator by a rope whose mass is negligible.What force does the mass exert on the rope when the elevator has an acceleration of 4.0 m/s2 upward?

The tension in a string from which a 4.0-kg object is suspended in an elevator is equal to 44 N.What is the acceleration of the elevator?

An object of unknown weight is suspended as shown.The tension in rope 1 is 25 lb,and the tension in rope 2 is 31 lb.What is the weight of the suspended object?

A 3.0-kg block slides on a frictionless 20° inclined plane.A force of 16 N acting parallel to the incline and up the incline is applied to the block.What is the acceleration of the block?

At an instant when a 4.0-kg object has an acceleration equal to M/s2,one of the two forces acting on the object is known to be n.Determine the magnitude of the other force acting on the object.

A block is pushed up a frictionless 30° incline by an applied force as shown.If F = 25 N and M = 3.0 kg,what is the magnitude of the resulting acceleration of the block?

A 5.0-kg mass is suspended by a string from the ceiling of an elevator that is moving upward with a speed which is decreasing at a constant rate of 2.0 m/s in each second.What is the tension in the string supporting the mass?

If M = 1.1 kg,what is the tension in string 1?

The horizontal surface on which the block slides is frictionless.If F = 20 N and M = 5.0 kg,what is the magnitude of the resulting acceleration of the block?

If the only forces acting on a 2.0-kg mass are N and N,what is the magnitude of the acceleration of the particle?

A person weighing 0.70 kN rides in an elevator that has an upward acceleration of 1.5 m/s2.What is the magnitude of the force of the elevator floor on the person?

If M = 2.0 kg,what is the tension in string 1?

A 5.0-kg object is suspended by a string from the ceiling of an elevator that is accelerating downward at a rate of 2.6 m/s2.What is the tension in the string?

The only two forces acting on a body have magnitudes of 20 N and 35 N and directions that differ by 80°.The resulting acceleration has a magnitude of 20 m/s².What is the mass of the body?

In the figure,if the tension in string 1 is 34 N and the tension in string 2 is 24 N,what is the mass of the object shown?

Two forces are the only forces acting on a 3.0-kg object which moves with an acceleration of 3.0 m/s² in the positive y direction.If one of the forces acts in the positive x direction and has a magnitude of 8.0 N,what is the magnitude of the other force?

A 5.0-kg mass is attached to the ceiling of an elevator by a rope whose mass is negligible.What force does the mass exert on the rope when the elevator has an acceleration of 4.0 m/s² upward?

At an instant when a 4.0-kg object has an acceleration equal to M/s²,one of the two forces acting on the object is known to be n.Determine the magnitude of the other force acting on the object.

A person weighing 0.70 kN rides in an elevator that has an upward acceleration of 1.5 m/s².What is the magnitude of the force of the elevator floor on the person?

A 5.0-kg object is suspended by a string from the ceiling of an elevator that is accelerating downward at a rate of 2.6 m/s².What is the tension in the string?