Physics Pearson 6.1 lesson check

  1. ^^State:^^ How is work calculated when force and displacement are in the same direction?

   Step 1: In this problem, we are asked to state how work is calculated when the force and displacement are in the same direction.

   Step 2: When F⃗F and d⃗d are parallel, the work done is simply the product of their magnitudes, that is

                                                      ==W=FdW=Fd​==

  1. ^^Identify:^^ Which component of force is used to calculate work when the force and displacement are at an angle to each other?

Step 1: In this problem, we must determine which component of the force is used to calculate work done when the force and displacement are at an angle.

Step 2: The component of the force is Fcos⁡θFcosθ, where θθ is the angle between the force and displacement. This component is the component parallel to the displacement.

  1. ^^Assess:^^ Is it possible to do work on an object that remains at rest? Explain.

Step 1: In this problem, we are asked if we can perform work on an object that is at rest.

Step 2: It is not possible. The amount of work done is proportional to the displacement, and an object at rest has zero displacement. The object has received no work.

  1. ^^Give Examples:^^ Give two examples of (a) positive work done by a frictional force (if possible) and (b) negative work done by a frictional force.

Step 1: It is not possible. The amount of work done is proportional to the displacement, which is zero for an object at rest. There has been no work done on the object.

Step 2: Part A

Static friction is the only force that can cause positive friction. Two blocks are stacked on top of each other, and the lower block is pulled with such force that the upper block does not slip. The static friction would cause the upper block to move in a direction parallel to the applied force and the displacement of the two blocks, resulting in positive work done.

Step 3: A child riding a sled is another example of static friction doing positive work. The sled's static friction on the child keeps the child moving forward. The sled's static friction has a positive effect on the child.

Step 4: Part B

Kinetic friction always does negative work. A car that turns off its engine, for example, will slow down because kinetic friction will apply a force opposite the direction of the car's velocity and displacement, doing negative work.

Step 5: A person sliding across a rough surface is another example. Because kinetic friction is always opposite displacement, the work done is negative.

  1. ^^Calculate:^^ A child in a tree house uses rope attached to a basket to lift a 22-N dog upward through a distance of 4.7 m into the house. How much work does the child do in lifting the dog?

Step 1: In this problem, we are told that a child in a tree house lifts a dog weighing w = 22 pounds.

N w=22 N and upward for d = 4.7

m\sd=4.7 m. We measure the amount of work done by the child.

Step 2: The force exerted by the child must be equal to the weight of the dog, so F=wF=w. We have

W=Fd

=we

=(22N)(4.7m)

=103.4 j

==W=1.0×102 J==

  1. ^^Calculate:^^ To move a suitcase up to the check-in stand at an airport, a student pushes with a horizontal force through a distance of 0.95 m. If the work done by the student is 32 j, what is the magnitude of the force he exerts?

Step 1: In this problem, a student moves a suitcase over a distance of d = 0.95.

m d=0.95 m, and the work completed is W = 32

J\sW=32 J. We compute the student's applied force.

Step 2: Force and displacement are both parallel. From the definition of work, we have.

w=fd

--→ f= w/d

=32 J/ 0.95 m

= 33.68421 N

==f= 34 N ​==

  1. ^^Calculate:^^ A farmhand pushes ​a bale of hay 3.9 m across the floor of a barn. she exerts a force of 88 N at an angle of 25 degrees below the horizontal. How much work has she done?

   Step 1: In this problem, a farmhand pushes a bale of hay over a displacement of d=3.9 md=3.9 m horizontally. She applies a force F=88 NF=88 N at an angle θ=25∘θ=25∘ below the horizontal. We calculate the work done by the farmhand on the hay.

   Step 2: The angle between the force and displacement is θθ. The work done must be

   W​=F⋅d

   = fd cos θ

   = (88 N)(3.9m) cos 25∘

   = 311.04483 J

   ==W=310 J==