Question 1

The work expended to accelerate a car from 0 to 30 m/s

Accepted Answer

The work done on an object to accelerate it is directly proportional to the change in velocity and the mass of the object. In this case, the change in velocity from 0 to 30 m/s is smaller than the change in velocity from 30 m/s to 60 m/s, therefore the work expended to accelerate the car from 0 to 30 m/s is less than that required to accelerate it from 30 m/s to 60 m/s.

Question 2

If mechanical work is done on a body,the body must

Accepted Answer

Mechanical work is defined as the product of force and displacement in the direction of force. Therefore, if mechanical work is done on a body, it means that a force has been applied and the body has moved in the direction of the force. Thus, the correct option is E- move.

Question 3

The kinetic energy of a car is 1.00 $\times$ 10⁵ J.If the car's speed is increased by 20 percent,the kinetic energy of the car becomes

Accepted Answer

The kinetic energy formula is KE=1/2mv^2, where m is the mass of the car and v is its speed. Since the mass of the car is not given and does not change, we can ignore it.

Using the given formula, we have KE1 = 1.00x10^15 J.

When the speed is increased by 20%, the new speed becomes v' = 1.2v. Therefore, the new kinetic energy is:
KE2 = 1/2m(1.2v)^2 = 1.44(1/2mv^2) = 1.44KE1

Thus the kinetic energy of the car becomes 1.44x10^15 J, which corresponds to choice C.

Question 4

A particle moves halfway around a circle of radius R.It is acted on by a radial force of magnitude F.The work done by the radial force is

Accepted Answer

The work done by a radial force is zero because the force is always perpendicular to the direction of motion, and work is calculated as the dot product of force and displacement, which is zero for perpendicular vectors.

Question 5

The average marathon runner can complete the 42.2-km distance of the marathon in 3 h and 30 min.If the runner's mass is 75 kg,what is the runner's average kinetic energy during the run?

Accepted Answer

The answer of The average marathon runner can complete the...

Question 6

Use the figure to answer the question. A light rope runs through two frictionless pulleys of negligible mass.A mass,m,is hung from one of the pulleys and a force,F is applied to one end of the rope so that the mass moves at a constant speed. 
-If the force is applied over a distance,d,the distance the mass moves is

Accepted Answer

The answer of Use the figure to answer the question....

Question 7

A body moves with decreasing speed.Which of the following statements is true?

Accepted Answer

When a body moves with decreasing speed, it means its kinetic energy is decreasing. According to the work-energy principle, the net work done on a body is equal to the change in its kinetic energy. Here, the kinetic energy is decreasing, therefore the net work done on the body must be negative. Hence, the correct option is (E) – The net work done on the body is negative, and the kinetic energy is decreasing.

Question 8

Negative work means

Accepted Answer

Negative work occurs when the applied force is in the opposite direction of the displacement of the object. This means that the work being done is taking away energy from the object, causing its kinetic energy to decrease. Choice A is incorrect because increasing kinetic energy would indicate positive work. Choice B is not a defining characteristic of negative work. Choice C describes the case where no work is being done, as the force and displacement are perpendicular. Choice E is not true, as negative work is a valid concept in physics.

Question 9

Use the figure to answer the question. A light rope runs through two frictionless pulleys of negligible mass.A mass,m,is hung from one of the pulleys and a force,F is applied to one end of the rope so that the mass moves at a constant speed. 
-A car and its occupants,with a total mass m,are moving at a speed v relative to an observer on the side of the road.You are driving in another car,also traveling at speed v,and in the same direction as the first car.The kinetic energy of the first car relative to you is

Accepted Answer

The kinetic energy of the first car relative to the observer on the side of the road is given by 1/2mv^2. However, since both cars are moving at the same speed v in the same direction, their relative speed is 0. Therefore, the kinetic energy of the first car relative to the observer is 1/2m(0)^2 = 0.

Question 10

Use the figure to answer the question. A light rope runs through two frictionless pulleys of negligible mass.A mass,m,is hung from one of the pulleys and a force,F is applied to one end of the rope so that the mass moves at a constant speed. 
-What is the force needed to move the mass at a constant speed?

Accepted Answer

According to Newton's second law of motion, force equals mass times acceleration (F=ma). In this case, since the mass is moving at a constant speed, its acceleration is zero. Therefore, the net force acting on the mass must also be zero. This means that the force applied to the rope (F) must be equal in magnitude and opposite in direction to the force of gravity acting on the mass (mg). Therefore, the force needed to move the mass at a constant speed is F = mg.

Question 11

A 5-kg object slides down a frictionless surface inclined at an angle of 30º from the horizontal.The total distance moved by the object along the plane is 10 meters.The work done on the object by the normal force of the surface is

Accepted Answer

Since the surface is frictionless, the only force acting on the object is its weight which acts down the plane. The component of the weight acting along the plane is given by:

w(sin30º) = (5 kg)(9.8 m/s^2)(0.5) = 24.5 N

Since the object is sliding down the plane, the displacement and the direction of the weight are the same. Therefore, the work done by the weight force on the object is given by:

W = Fd = (24.5 N)(10 m) = 245 J

However, the normal force of the surface is acting perpendicular to the displacement, so the work done by this force is zero.

Thus, the work done on the object by the normal force of the surface is zero, so the answer is A) zero.

Question 12

The weight of an object on the moon is one-sixth its weight on Earth.A body moving with a given speed on the moon has kinetic energy equal to ______________ it would have if it were moving at the same speed on Earth.

Accepted Answer

The kinetic energy of an object is given by the formula $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the object and $v$ is its velocity. This formula shows that kinetic energy is independent of the gravitational field in which the object is located, depending only on its mass and speed. Therefore, a body moving at a given speed on the moon has the same kinetic energy as it would if it were moving at the same speed on Earth.

Question 13

A constant force of 45 N directed at angle $\theta$ to the horizontal pulls a crate of weight 100 N from one end of a room to another a distance of 4 m.Given that the vertical component of the pulling force is 12 N,calculate the work done by the force in moving the crate.

Accepted Answer

The answer of A constant force of 45 N directed...

Question 14

A bullet with a mass of 12 g moving horizontally strikes a fixed block of wood and penetrates a distance of 5.2 cm.The speed of the bullet just before the collision is 640 m/s.The average force that the wood exerted on the bullet was

Accepted Answer

The average force can be calculated using the work-energy principle. The work done by the force is equal to the change in kinetic energy of the bullet. The initial kinetic energy is (1/2)mv^2, and the final kinetic energy is 0. The work done is also equal to force times distance. Solving for force gives approximately 4.7 x 10^4 N.

Question 15

Initially a body moves in one direction and has kinetic energy K.Then it moves in the opposite direction with three times its initial speed.What is the kinetic energy now?

Accepted Answer

Kinetic energy is proportional to the velocity squared. When the body moves in the opposite direction with three times its initial speed, its velocity becomes -3v (where v is the initial velocity). Therefore, the kinetic energy becomes (1/2)mv^2 * 9 = 9K.

Question 16

Car drag racing takes place over a distance of a  mile (402 m)from a standing start.If a car (mass 1500 kg)could be propelled forward with a pulling force equal to that of gravity,what would be the change in kinetic energy and the terminal speed of the car (in mph)at the end of the race be? (For comparison,a modern,high-performance sports car may reach a terminal speed of just over 100 mph = 44.7 m/s.)

Accepted Answer

The answer of Car drag racing takes place over a...

Question 17

The SI unit of energy can be expressed as

Accepted Answer

The SI unit of energy is the joule (J), which is not correctly represented by any of the options given. A joule is defined as the amount of work done when a force of one newton is applied over a distance of one meter, which can be expressed as kg·m²/s².

Question 18

The work-kinetic energy theorem states that

Accepted Answer

The work-kinetic energy theorem states that the work done on an object is converted into kinetic energy and vice versa. This means that when work is done on an object, its energy increases as kinetic energy, and when an object's kinetic energy decreases, it is converted into work done on the object. Therefore, choice A is the correct answer.

Question 19

A force pushes a block along a horizontal surface against a force of friction _f.If the block undergoes a displacement s at constant velocity,the work done by the resultant force on the block

Accepted Answer

When an object moves with constant velocity, its kinetic energy does not change. This means that the work done by the resultant force is equal and opposite to the work done by the force of friction, and the net work done on the object is zero. Therefore, the correct answer is E.

Question 20

Two forces,both of magnitude 12 N and directed 35 degrees either side of the direction of motion,pull a crate 15 m.How much work is done by the forces in moving the crate?

Accepted Answer

The resultant force in the direction of motion can be found by adding the horizontal components of the two forces. Each force has a horizontal component of $12 \cos(35^\circ)$ N. Therefore, the total force in the direction of motion is $2 \times 12 \cos(35^\circ)$ N. Work done is the product of this force and the distance (15 m) moved in the direction of the force. Thus, the work done is $2 \times 12 \cos(35^\circ) \times 15$ J, which calculates to approximately 2.1 kJ, making option E the correct answer.

Quiz 6: Work and Energy