Chapter6 Tutorial physics

Tutorial-9 Chapter 6: Work and Kinetic Energy Date: 22-04-2012

Q-1). An object moved under the action of an applied force F⃗= [(2x) i �̂ +(4y2)j ̂� ] N, where x and y are in meters. If the object is displaced from an initial position ¿ ) m to a final position (4 i+7 j )m, how much work is done by the force during this displacement ?

Q-2). A block of ice with mass 2.00 Kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9o below the horizontal. If the block of ice starts from rest, what is its final speed ? Ignore friction. (YF-13Ed-Q6.28)

Q-3). A child applies a force F⃗ parallel to the x-axis to a 10.0 kg sled moving on the frozen surface of a small pond. As the child controls the speed of the sled, the x-component of the force she applied varies with the x-coordinate of the sled as shown in the figure. Calculate the work done by the force F⃗ when the sled moves (YF-13Ed-Q6.34 & 6.35)

(a) from x=0 to x =8.0 m

(b) from x =8.0 m to x= 12.0 m


(c ). From x =0 to 12.0 m

(d). Suppose the sled is initially at rest at x= 0. Use the work-energy theorem to find the speed of the sled at

(i) x =8.0 m (ignore friction) (ii) x = 12.0 m. (ignore friction)

Q-4). A 2.0-kg and a 3.0-kg box on a perfectly smooth horizontal floor have a spring of force constant 250 N/m compressed between them. If the initial compression of the spring is 6.0 cm, find the acceleration of each box the instant after they are released. Be sure to include free body diagrams of each box as part of your solution. (YF-13Ed-Q6.36)

Q-5). A 4.00 kg block of ice is placed against a horizontal spring that has force constant k = 200 N/m and is compressed 0.025 m. The spring is released and accelerates the block along a horizontal surface. You can ignore friction and the mass of the spring. (YF-13Ed-Q6.40)

(a) Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length.

(b). What is the speed of the block after it leaves the spring ?


Q-6). An ingenious bricklayer builds a device for shooting brick up to the top of the wall where he is working. He places a brick on a vertical compressed spring with force constant k = 450 N/m and negligible mass. When the spring is released the brick is propelled upward. If the brick has mass 1.80 kg and is to reach a maximum height of 3.6 m above its initial position on the compressed spring, what distance must the bricklayer compress the spring initially ? (YF-13Ed-Q6.46)

Q-7). A 20.0 kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200. What average power is produced by friction as the rock stops ? (YF-13Ed-Q6.52)


Q-8). An elevator has mass 600 kg, not including passengers. The elevator is designed to ascend , at constant speed, a vertical distance of 20.0 m in 16.0 s, and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator ? Assume that an average passenger has mass 65.0 kg. (YF-13Ed-Q6.56)

Q-1). 6.14. IDENTIFY: The book changes its speed and hence its kinetic energy, so work must have been done on it.

SET UP: Use the work-kinetic energy theorem with In part

(a) use and to calculate W. In parts (b) and (c) use and W to calculate EXECUTE: (a) Substituting the notation and

(b) Noting and

so

(c) Similarly, and


EVALUATE: Negative corresponds to a decrease in kinetic energy (slowing

down) and positive corresponds to an increase in kinetic energy (speeding up).

Q-2). 6.28. IDENTIFY: Apply

SET UP: The normal force does no work. The work W done by gravity is , where is the vertical distance the block has dropped when it has

traveled a distance L down the incline and is the angle the plane makes with the horizontal.

EXECUTE: The work-energy theorem gives . Using the

given numbers, EVALUATE: The final speed of the block is the same as if it had been dropped from a

height h.

Q-3(i) 6.34. IDENTIFY: The magnitude of the work can be found by finding the area under the graph.

SET UP: The area under each triangle is so the work done is positive when x increases during the displacement.EXECUTE: (a) (b) (c) EVALUATE: The sum of the answers to parts (a) and (b) equals the answer to part (c).

Q-3(ii) 6.35. IDENTIFY: Use the work-energy theorem and the results of Problem 6.30.

SET UP: For to For to

EXECUTE: (a)

(b) EVALUATE: is always in the For this motion does positive work and the speed continually increases during the motion.


Q-4). 6.36. IDENTIFY: The force of the spring is the same on each box, but they have different accelerations because their masses are different. Hooke’s law gives the spring force.SET UP: The free-body diagrams for the boxes are shown in Figure 6.36. Label the

boxes A and B, with and is the spring force and is the same for each box. We apply to each box.

Figure 6.36

EXECUTE:

The accelerations are in opposite directions.EVALUATE: The same magnitude of force is exerted on each object, but the acceleration that is produced by this force is larger for the object of smaller mass.

Q-5). 6.40. IDENTIFY: For the spring, Apply

SET UP: and

EXECUTE: (a)

(b) The work-energy theorem gives EVALUATE: The block moves in the direction of the spring force, the spring does positive work and the kinetic energy of the block increases.

Q-6). 6.46. IDENTIFY: Apply to the brick. Work is done by the spring force and by gravity.

SET UP: At the maximum height, Gravity does negative work, The

work done by the spring is where d is the distance the spring is compressed initially.EXECUTE: The initial and final kinetic energies of the brick are both zero, so the net

work done on the brick by the spring and gravity is zero, so or


The spring will provide an upward force while the spring and the brick are in contact. When this force goes to zero, the spring is at its uncompressed length. But when the spring reaches its uncompressed length the brick has an upward velocity and leaves the spring.EVALUATE: Gravity does negative work because the gravity force is downward and the brick moves upward. The spring force does positive work on the brick because the spring force is upward and the brick moves upward.

Q-7). 6.52. IDENTIFY: The thermal energy is produced as a result of the force of friction,

The average thermal power is thus the average rate of work done by

friction or

SET UP:

EXECUTE: EVALUATE: The power could also be determined as the rate of change of kinetic

energy, where the time is calculated from and a is calculated from a

force balance,

Q-8)6.56. IDENTIFY and SET UP: Use Eq. (6.15) to relate the power provided and the amount of work done against gravity in 16.0 s. The work done against gravity depends on the total weight which depends on the number of passengers.EXECUTE: Find the total mass that can be lifted:

so

This is the total mass of elevator plus passengers. The mass of the passengers is

The number of passengers is 28 passengers can ride.

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