1. A man pushes on a box at an angle of 30.0° with a force of 20.0 N and the box moves across the floor. Which of the following component of the force do you use to calculate work?
(Points : 1)
0.00 N
10.0 N
17.3 N
20.0 N
Question 2.2. A man carries a 10 kg sack of groceries in his arms with a force of 50 N as he walks forward a distance of 10 m. How much work has he done?
(Points : 1)
0 J
50 J
100 J
500 J
Question 3.3. In which of the following situations has the energy of the object been increased?
(Points : 1)
A hammer thrower spins around while preparing to release the hammer.
A horse moves horizontally while a circus performer jumps vertically on its back.
A boy pushes a lawn mower forward by pushing at a downward angle on the handle.
A soldier runs forward, while holding a rifle above his head.
Question 4.4. A rope pulls on a metal box at an angle of 60.0° with a force of 255 N. The box moves horizontally for 15.0 m. How much work was done on the box?
(Points : 1)
255 J
1,910 J
3,310 J
3,830 J
Question 5.5. Which of the following is a scalar quantity?
(Points : 1)
parallel component of force
perpendicular component of force
displacement
work
1. 17.3 N
2. 0 J
3. A boy pushes a lawn mower forward by pushing at a downward angle on the handle
4. 1,910 J
5. Work
(I just took the test, these are the correct answers)
Emily L is correct
To answer these questions, we need to understand some concepts related to work and energy.
1. When calculating work, we use the component of force that is parallel to the displacement. In this case, the force is applied at an angle of 30.0°. To find the parallel component of the force, we use the equation F_parallel = F * cos(theta), where F is the magnitude of the force (20.0 N) and theta is the angle (30.0°). Plugging in the numbers, we get F_parallel = 20.0 N * cos(30.0°) ≈ 17.3 N.
2. Work is defined as the product of force and displacement. In this case, the man applies a force of 50 N and moves the sack forward a distance of 10 m. The work done can be calculated using the equation W = F * d, where W is the work done, F is the force (50 N), and d is the displacement (10 m). Plugging in the numbers, we get W = 50 N * 10 m = 500 J.
3. In this question, we're asked about situations in which the energy of the object has been increased. Energy can be increased by doing work on an object. Looking at the options:
- A hammer thrower spinning around is not doing work on the hammer, so the energy is not increased.
- A horse moving horizontally while a circus performer jumps vertically on its back does not involve any work on the horse, so the energy is not increased.
- A boy pushing a lawn mower forward by pushing at a downward angle on the handle is doing work against the force of friction, so the energy of the mower is increased.
- A soldier running forward while holding a rifle above his head does not involve any work on the rifle, so the energy is not increased.
4. To determine the work done on the box, we need to calculate the component of the force that is parallel to the displacement. The force applied at an angle of 60.0° is 255 N. Using the equation F_parallel = F * cos(theta), where F is the magnitude of the force (255 N) and theta is the angle (60.0°), the parallel component is F_parallel = 255 N * cos(60.0°) ≈ 127.5 N. To find the work done, we multiply this parallel force by the horizontal displacement, so W = F_parallel * d = 127.5 N * 15.0 m = 1910 J.
5. Work is a scalar quantity, meaning it only has magnitude and no direction. The options provided include the parallel component of force, perpendicular component of force, displacement, and work. Only work is a scalar quantity, as it represents the transfer of energy without considering any specific direction.
1. Fx = 20*cos30 = 17.32 N.
2. Work = Fx*d = 0 * 10 = 0
3. a.
4. W = Fx*d = 255*Cos60 * 15 =
5.