Question

14. A block rests on a frictionless table on Earth. After a 20-N horizontal force is applied to the block, it accelerates at 3.9 m/s2. Then the block and table are carried to the moon, where the acceleration due to gravity is 1.62 m/s2. A horizontal force of 10 N is then applied to the block. What is the acceleration?
A. 1.8 m/s2
B. 2.5 m/s2
C. 2.1 m/s2
D. 2.3 m/s2
E. 2.0 m/s2

15. Workers are trying to free an SUV stuck in the mud. To move the vehicle, they use three ropes held parallel to the ground, producing the force vectors shown in the figure. What is the magnitude and direction of the resultant force of the three pulls?

A. 2160 N at 35.5° counterclockwise to the x-axis
B. 16 50 N at 65.9° counterclockwise to the x-axis
C. 866 N at 78.1° counterclockwise to the x-axis
D. 1510 N at 83.2° counterclockwise to the x-axis

16. A large sphere has a mass of 175 kg and is suspended by a chain from the ceiling. The mass of the chain is 12.0 kg. What is the tension in the chain?

A. 1940 N
B. 1830 N
C. 1610 N
D. 1720 N

17. Two chains are available to hold up an object. A student arranges the object and either one or two chains according to the diagrams shown here. Which arrangement generates the most tension in one chain?

A. B
B. D
C. A
D. C

18. A force of gravity between the sun and its planets holds the planets in orbit around the sun. If that force of gravity suddenly disappeared, in what kind of path would the planets move?

19.
A car of mass 1200 kg is traveling at 2 m/s. Then it collides with a heavy boulder in the road. During the collision, the car moves forward only 15 cm before stopping. What is the average force exerted by the boulder on the car?

A. 1.8 × 104 N
B. 1.6 × 104 N
C. 5.4 × 104 N
D. 3.2 × 104 N
E. 6.5 × 105 N

20. Which type of force can act through empty space?

A. magnetism
B. normal
C. spring
D. tension

21. A 5.0-kg block and a 4.0-kg block are connected vertically by a 0.6-kg rod. In the diagram, the links between the blocks and the rod are denoted by the letters A and B. A force ->F is applied to lift the upper block. The blocks and rod assembly move upward at constant velocity.
What is the magnitude of the force in link A?

A. 41 N
B. 39 N
C. 47 N
D. 43 N
E. 45 N

22. At which location would a bowling ball have the greatest weight?

A. on Earth, several kilometers below the surface
B. on the surface of the moon
C. on Earth at sea level
D. on Earth, at the top of a tall mountain

23. A worker with spikes on his shoes applies a constant horizontal force of 20 N to push a 40-kg box across a frictionless, frozen lake. What is the acceleration of the box?

A. 0.5 m/s2
B. 2.0 m/s2
C. 8.0 m/s2
D. 4.1 m/s2

24. A batter swings a bat and hits a baseball into the outfield. During the collision between bat and ball, how do the forces on the two objects compare?

A. The forces are opposite in direction, with the force on the ball much stronger in magnitude.
B. The forces are opposite in direction, with the force on the bat much stronger in magnitude.
C. Both forces are equal in magnitude and direction.
D. Both forces are equal in magnitude but opposite in direction.

25. A superhero throws a 2400-N boulder at an adversary. What horizontal force must the superhero apply to the boulder to provide it with a horizontal accelertion of 12.0 m/s2?

A. 245 N
B. 2940 N
C. 28800 N
D. 200 N

26. The acceleration due to gravity on the surface of Mercury is 3.71 m/s2. An object with a mass of 8.69 kg has what weight on Mercury?

A. 23.4 N
B. 85.2 N
C. 42.7 N
D. 32.2 N

Answer 1

14. E 2.0 m/s^2

Initially, a 20 N force is applied to the block, so it has an acceleration of 3.9 m/s^2. According to Newton's law, the mass of the block is:

$m=\frac{F}{a}=\frac{20 N}{3.9 m/s^2}=5.13 kg$

In the second situation, a force of 10 N is applied to the block. Since the mass is still the same, the acceleration now is:

$a=\frac{F}{m}=\frac{10 N}{5.13 kg}=1.95 m/s^2$

So, approximately 2.0 m/s^2.

15. C. 866 N at 78.1° counterclockwise to the x-axis

Resultant along the x- and y-axis:

$R_x = (985 N)(cos 31^{\circ})-(788 N)(sin 32^{\circ})-(411 N)(cos 53^{\circ})=179.4 N$

$R_y = (985 N)(sin 31^{\circ})+(788 N)(cos 32^{\circ})-(411 N)(sin 53^{\circ})=847.3 N$

Magnitude and direction of the resultant:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(179.4 N)^2+(847.3 N)^2}=866.0 N$

$\theta=arctan(\frac{R_y}{R_x})=\arctan(\frac{847.3 N}{179.4 N})=78.1^{\circ}$

16. D. 1720 N

Since the sphere is suspended, it is in equilibrium, therefore the tension in the chain is equal to the weight of the sphere attached to it, therefore:

$T=mg=(175 kg)(9.81 m/s^2)=1720 N$

17. C. A

This arrangement generates the largest tension in the chain, because in all other arrangements the weight of the object is split between the two chains, while in this case all the weight is hold by one chain, therefore the tension in this case is larger.

18. Straight path

The gravity "holds" the planets keeping them in a circular orbit. If we remove gravity, the planets would continue in a straight path with constant speed, because now there are no more forces acting on it, so by inertia they will continue their uniform motion with constant speed.

19. B. 1.6 × 104 N

First we can find the deceleration of the car by using the SUVAT equation:

$v^2 -u^2 =2aS$

where v=0 m/s, u=2 m/s, and S=15 cm=0.15 m. Re-arranging, we have

$a=\frac{-u^2}{2S}=\frac{-(2 m/s)^2}{2(0.15 m)}=-13.3 m/s^2$

And now we can calculate the average force exerted on the car, by using Newton's second law:

$F=ma=(1200 kg)(-13.3 m/s^2)=-15960 N=-1.6 \cdot 10^4 N$

(the negative sign means that the force's direction is opposite to the motion of the car)

20. A. magnetism

Magnetism is part of the electromagnetic force, which is one of the fundamental forces which act also through empty space. All the other forces need some object in order to act.

21. E. 45 N

The magnitude of the force in link A is equal to the weight of the rod plus the weight of the lower block, therefore:

$W=(m_1 + m_2)g=(0.6 kg+4.0 kg)(9.8 m/s^2)=45 N$

22. C. on Earth at sea level

The weight of the bowling ball is given by: $W=mg$, where m is the mass of the ball and g is the acceleration due to gravity. The value of g increases when moving from the Earth's center to the Earth's surface, then decreases when moving far from the surface, so the point where g is greatest is at sea level, where it is 9.81 m/s^2. On the surface of the Moon, g is much smaller (about 1/6 of the value on Earth).

23. A. 0.5 m/s2

The acceleration of the block is given by Newton's second law:

$a=\frac{F}{m}=\frac{20 N}{40 kg}=0.5 m/s^2$

24. D. Both forces are equal in magnitude but opposite in direction.

According to Newton's third law: if an object A exerts a force on an object B, then object B exerts a force equal and opposite on object B. In this case, objects A and B are the bat and the baseball, therefore the two forces are equal in magnitude and opposite in direction.

25. B. 2940 N

The mass of the boulder is equal to its weight divided by the acceleration of gravity (9.81 m/s^2):

$m=\frac{W}{g}=\frac{2400 N}{9.81 m/s^2}=245 kg$

So now we can calculate the force needed to accelerate the boulder to 12.0 m/s^2:

$F=ma=(245 kg)(12.0 m/s^2)=2940 N$

26. D. 32.2 N

The weight of the object on Mercury is given by:

$W=mg=(8.69 kg)(3.71 m/s^2)=32.2 N$

Answer 2

14. Let's find mass. We know that F = m*a, so m = F/a.m = 20/3,9 m/s^2 = 5,12 kg;We already know the mass, so acceleration will be a = F/m;a = 10/5,12 = 2,0 m/s^2 (approximately);Answer: E.
15. According to the picture given above, let's define all components:These are refer to y: $985N sin(31)= 507;788N cos(32)= 668;411N sin(53)=-328;$
And these are refer to x: $985N cos(31)= 844 788N sin(32)= -417 411N cos(53)= -247$
And finally let's calculate tan:  $tan^-1 (507/844)= 30,98; tan^-1 (668/-417)= -58,02; tan^-1 (-328/-247)= 53,01$
According to these calculations we've got: Sum Fx = 179,4 and Sum Fy = 847. Then let's count magnitude:$Fsum = rad 179^2 + 847^2 = 865$
Then we've got [tex]cos^-1 (179/865) = 77.7 (approximately).
So the most approximate answer is C. 866 N at 78.1° counterclockwise to the x-axis.
16. I think that this question is incomplete because there wasn't mentioned in what end of the chain the tension should be calculated. Anyway I'll help you with both bottom and top ends. We will use the basic formula W=m*g; W(bottom) = 175*9.8 = 1715 N; W(top) = (175+12)*9.8 = 1833 N; So you should choose between B. 1830 N or D. 1720 N. But I think the most possible answer is B.
17. I am definitely sure that A diagram generates the most tension in one chain. So the answer is C. The box is held by 1 chain and have all its weight.
18. I think that each planet would move in a straight line at constant speed, because there will be a zero gravity condition and there won't any impact on the planets.
19.  According to the Work-Energy Theorem we have:1/2*1200*2^2 = 2400 (J);Then let's count the average force according to the formula: Work =  F*D where D is displacement.F = 2400/ 0.15 = 16000 (N)
So the answer is B. 1.6 × 10^4 N
20. If my memory serves me well, magnetism is the force that can act through empty space. It's one aspect of the combined electromagnetic force. So the answer is A.
21. According to the illustration given above, I think we should use formula F = m*g; Let's count m. m = 5 kg - 0,6 kg = 4,4 kg; Then we have everything to count force:F = 4,4 * 9,8 = 43,1 N. So the most approximate answer is D. 43 N.
22. I am definitely sure that the answer is C. on Earth at sea level. Because weight has the formula W=mg. And on the earth surface the magnitude of g is higher that everywhere so the greatest weight is on Earth at sea level.
23. We have everything to calculate the acceleration. According to Newton's second law, the formula is a = F/m;a = 20/40 = 0,5 m/s^2; So the answer is A.
24. According to the definition of action reaction forces, the answer should be: A. The forces are opposite in direction, with the force on the ball much stronger in magnitude.
25. I am pretty sure that we should use Newton’s second law of motion F = m*a. First we should find mass, using formula W=m*g => m = W/g => m = 2400/9.8= 245 kg. Then we can find F.F = m*a;F = 245*12 = 2940 N; Answer: B