Question 1

An object moves in a circle at constant speed. The work done by the centripetal force is zero because:

Accepted Answer

The work done by a force is zero if the force is perpendicular to the direction of displacement. In circular motion, the centripetal force always acts perpendicular to the velocity of the object, hence no work is done by the centripetal force.

Question 2

Which of the following is the correct combination of dimensions for energy?

Accepted Answer

C

Question 3

A boy holds a 40-N weight at arm's length for 10 s. His arm is 1.5 m above the ground. The work done by the force of the boy on the weight while he is holding it is:

Accepted Answer

The work done by a force is given by the equation W = Fd cosθ, where F is the force applied, d is the displacement of the object, and θ is the angle between the force and the displacement. In this case, the force of the boy on the weight is equal to the weight of the object, which is 40 N. The displacement of the object is zero, since it is held at a constant height for 10 s. Therefore, the angle between the force and the displacement is 90 degrees, and cosθ = 0.

Putting these values into the equation, we get W = Fd cosθ = 40 N × 0 m × 0 = 0 J. Therefore, the work done by the force of the boy on the weight while he is holding it is 0 J.

Question 4

A sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting a horizontal force on it. The coefficient of kinetic friction between sledge and snow is 0.05. How much work is done by the dog team pulling the sledge 1000 m at constant speed?

Accepted Answer

The answer of A sledge (including load) weighs 5000 N....

Question 5

Camping equipment weighing 6000 N is pulled across a frozen lake by means of a horizontal rope. The coefficient of kinetic friction is 0.05. How much work is done by the campers in pulling the equipment 1000 m if its speed is increasing at the constant rate of 0.20 m/s²?

Accepted Answer

First, we need to find the magnitude of the force required to pull the camping equipment at a constant speed. We know that the force of friction is given by:

F_friction = coefficient of friction * F_normal

where F_normal is the normal force acting on the equipment, which is equal to its weight in this case (since it is on a horizontal surface). Therefore,

F_friction = 0.05 * 6000 N = 300 N

Since the equipment is moving at a constant speed, the force pulling it must be equal and opposite to the force of friction, which is 300 N in this case.

Now, we can calculate the work done by the campers in pulling the equipment 1000 m with a speedup of 0.20 m/s². We know that the work done is equal to the change in kinetic energy of the equipment, given by:

W = (1/2) * m * (v_f² - v_i²)

where m is the mass of the equipment (which we can find by dividing the weight by the acceleration due to gravity), v_f is the final velocity of the equipment after being pulled 1000 m, and v_i is the initial velocity (which we take to be zero).

m = 6000 N / 9.81 m/s² = 611.8 kg

v_f = (2*a*d + v_i²)^(1/2) = (2*0.20 m/s²*1000 m)^(1/2) = 20 m/s

W = (1/2) * 611.8 kg * (20 m/s)² = 244720 J

Therefore, the answer is D) 4.2 × 10^5 J (rounded to three significant figures).

Question 6

A particle moves 5 m in the positive x direction while being acted upon by a constant force  The work done on the particle by this force is:

Accepted Answer

The answer of A particle moves 5 m in the...

Question 7

Two trailers, X with mass 500 kg and Y with mass 2000 kg, are being pulled at the same speed. The ratio of the kinetic energy of Y to that of X is:

Accepted Answer

The kinetic energy of an object is given by the formula KE = 0.5 * m * v^2, where m is the mass of the object and v is its velocity. As both trailers are being pulled at the same speed, their velocities are equal. Therefore, the ratio of their kinetic energies will depend only on their masses.

Using the formula above, the kinetic energy of trailer X is KE_X = 0.5 * 500 * v^2 = 250v^2, and the kinetic energy of trailer Y is KE_Y = 0.5 * 2000 * v^2 = 1000v^2. Therefore, the ratio of their kinetic energies is KE_Y/KE_X = 1000v^2/250v^2 = 4:1.

Therefore, the correct answer is C) 4:1.

Question 8

An object of mass 1 g is whirled in a horizontal circle of radius 0.5 m at a constant speed of 2 m/s. The work done on the object during one revolution is:

Accepted Answer

The work done on an object is given by the formula W = Fd, where W is work, F is force, and d is displacement. In this case, the force acting on the object is the tension in the string that is keeping it moving in a circle. Since the string is always perpendicular to the direction of motion, the force does no work on the object (work done by a force is given by the dot product of the force and displacement vectors). Therefore, the work done on the object during one revolution is zero.

Question 9

Camping equipment weighing 6000 N is pulled across a frozen lake by means of a horizontal rope. The coefficient of kinetic friction is 0.05. The work done by the campers in pulling the equipment 1000 m at constant velocity is:

Accepted Answer

The work done by the campers in pulling the equipment can be determined by the equation:
Work = Force x Distance x cos(theta)
where Force is the pulling force, Distance is the distance traveled, and theta is the angle between the force and the displacement (since it is horizontal, cos(theta) = 1).

The pulling force can be determined using the equation:
Force = friction force
friction force = coefficient of friction x normal force
normal force = weight of the equipment = 6000 N

friction force = 0.05 x 6000 = 300 N

Therefore, the force applied by the campers is 300 N.

Plugging in the given values, we get:
Work = 300 x 1000 x 1 = 300000 J

Rounding to the nearest tenth, we get 300000 J = 3.0 x 10^5 J = 3.0 x 10^15 Nm = 3.0 x 10^15 J. Therefore, the best choice is C.

Question 10

Which of the following groups does NOT contain a scalar quantity?

Accepted Answer

Displacement, acceleration, and force are vector quantities, meaning they have both magnitude and direction.

Question 11

A man pulls a sled along a rough horizontal surface by applying a constant force  at an angle $\theta$above the horizontal. In pulling the sled a horizontal distance d, the work done by the man is:

Accepted Answer

The work done by a constant force is given by W = Fd cosθ, where F is the magnitude of the force, d is the displacement, and θ is the angle between the force and the displacement. In this case, the force is applied at an angle θ = F1F1F1S1F1F1F10 above the horizontal. Therefore, the work done by the man is W = (11eaafba_af47_61a8_a7a5_c54e5db7e10b_TB3976_11)(d)cos F1F1F1S1F1F1F10. The correct choice is B.

Question 12

The weight of an object on the moon is one-sixth of its weight on the Earth. The ratio of the kinetic energy of a body on the Earth moving with speed V to that of the same body moving with speed V on the moon is:

Accepted Answer

The kinetic energy of a body is given by the formula $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the body and $v$ is its velocity. This formula shows that kinetic energy is independent of the location of the body (whether on Earth or the moon) as it only depends on the mass and velocity of the body. Since the mass and velocity of the body remain the same in both locations, the ratio of the kinetic energies is 1:1.

Question 13

Which of the following bodies has the largest kinetic energy?

Accepted Answer

The kinetic energy is given by the equation KE=1/2mv^2. Therefore, the kinetic energy of an object depends on both its mass and speed. In this case, the object with the largest kinetic energy will be the one with the highest product of mass and speed. Choice C has the highest product of mass (2M) and speed (3V), and therefore has the largest kinetic energy.

Question 14

A 8000-N car is traveling at 12 m/s along a horizontal road when the brakes are applied. The car skids to a stop in 4.0 s. How much kinetic energy does the car lose in this time?

Accepted Answer

The answer of A 8000-N car is traveling at 12...

Question 15

Which of the following is NOT a correct unit for work?

Accepted Answer

Work is defined as the product of force and displacement in the direction of the force, and its SI unit is the joule. Erg, joule, and newton-meter are all correct units for work, reflecting different systems of measurement (CGS, SI, and the metric system, respectively). The option with "watt" is incorrect because a watt is a unit of power, not work. Power is the rate at which work is done or energy is transferred.

Question 16

A 2-kg object is moving at 3 m/s. A 4-N force is applied in the direction of motion and then removed after the object has traveled an additional 5 m. The work done by this force is:

Accepted Answer

The answer of A 2-kg object is moving at 3...

Question 17

A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external force of 3.0 N, always tangent to the track, causes the object to speed up as it goes around. The work done by the external force as the mass makes one revolution is:

Accepted Answer

The answer of A 0.50-kg object moves in a horizontal...

Question 18

A man wishes to pull a crate 15 m across a rough floor by exerting a force of 100 N. The coefficient of kinetic friction is 0.25. For the man to do the least work, the angle between the force and the horizontal should be:

Accepted Answer

The answer of A man wishes to pull a crate...

Question 19

A crate moves 10 m to the right on a horizontal surface as a woman pulls on it with a 10-N force. Rank the situations shown below according to the work done by her force, least to greatest.

Accepted Answer

The answer of A crate moves 10 m to the...

Question 20

An object is constrained by a cord to move in a circular path of radius 0.5m on a horizontal frictionless surface. The cord will break if its tension exceeds 16N. The maximum kinetic energy of the object can have is:

Accepted Answer

The answer of An object is constrained by a cord...

Quiz 7: Kinetic Energy and Work