The text is a physics problem involving an Amazon delivery person driving a truck up a steep hill with an incline of 26°. A 3.5 kg box slides 2.1 m along the truck floor to the rear door due to the incline and movement. The coefficient of kinetic friction between the box and the truck floor is 0.36. Here's a breakdown of the calculations required:1. **Work Done by the Weight of the Box (Wmg)**: - Calculated as 31.6 J (joules).2. **Work Done by the Normal Force (WN)**: - This value is 0 J, as the normal force does no work.3. **Work Done by the Frictional Force (Wfk)**: - Calculated as 23.31 J. This is incorrect according to the text.4. **Net Work Done on the Box (Wnet)**: - The problem asks for this value to be calculated in joules.5. **Change in Kinetic Energy of the Box (ΔKE)**: - The problem asks for this value in joules.6. **Final Speed of the Box (vfinal)**: - The problem asks for this speed in meters per second (m/s).The task involves applying concepts of work and energy, including the effects of friction, incline, and gravitational force on the box.

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Description: The text is a physics problem involving an Amazon delivery person driving a truck up a steep hill with an incline of 26°. A 3.5 kg box slides 2.1 m along the truck floor to the rear door due to the incline and movement. The coefficient of kinetic fr

Given:- The mass of the box is m = 3.5 kg. the box slide direction along the floor to the rear door is d = 2.1 m. The person is driving up a steep hill with an incline of θ= 26°. The coefficient of kinetic friction between the box and the floor is μ= 0.36. Find:- a) What is the work done by the weight of the box? b) What is the work done by the normal force? c) What is the work done by the friction force? d) What is the net work done on the box? e) What is the change in kinetic energy o the box? f) How fast is the box moving just before it hits the rear door? Solution:- a) We know to draw a free-body diagram of the given system of the box which is inclined with the angle of θ= 26°as So now the free body diagram of the system is given as We know that the work done by the gravity force is given as Wg= mgsinθ×dWg = 3.5×9.8sin26×2.1Wg = 31.5758≈31.6Wg= 31.6 J This is the work done by gravity. b) We know that the normal force act perpendicular to the plane and we know that the displacement is in the direction of the plane. So the angle between the force and displacement is 900 So the work done by the normal force is given as WN= Ndcosϕ here the ϕ is an angle between the displacement and the force [normal force ].= 90o So now WN= Ndcos90WN =0 J This is the work done by the normal force. c) We know that the work done by the friction force is given as Wfr=frdcosϕ here the angle ϕ is 1800 because this force act in the opposite direction of the force of friction. So now Wfr = μNdcos180Wfr = -μmgcosθ×dWfr = -0.36×3.5×9.8cos26×2.1Wfr = -23.306 J ≈-23.31Wfr =-23.31 J This is work done by the force of friction on the box. d) The net work done is given as Wnet = Wg+Wfr+WNWnet = 31.6-23.306+0Wnet =8.2935≈8.3Wnet= 8.3 J This is the net force on the box. e) We know that the work-energy theorem, according to that the net work done on an object is equal to the change in the kinetic energy. Wnet=∆K.E∆K.E = 8.3 J This is the change in the kinetic energy of the object. f) We use the work-energy theorem here as Wnet=∆K.E12mv2-12mu2 = Wnet12×3.5v2-0= 8.3v = 8.3×23.5v = 2.1778≈2.2v = 2.2 m/s This is the velocity with the box moving just before it hits the rear door.a) The work done by the weight of the box is 31.6 J. b) The work done by the normal force is 0 j. c) The work done by the force of friction is -23.31 J. d) The work done by the net force is 8.3 J. e) The change in kinetic energy is 8.3 J. f) The velocity of the box moving just before it hits the rear door is 2.2 m/s.

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The overworked Amazon delivery person is driving up a steep hill with an incline of 26° when a box they forgot to secure starts sliding toward the back of the truck. The 3.5 kg box starts from rest near the drivers seat and slides 2.1 m along the floor to the rear door. The coefficient of kinetic friction between the box and the floor is 0.36. What is the work done by the weight of the box? Wmg = 31.6 What is the work done by the Normal Force? WN = 0 What is the work done by the frictional force? W₁ = 23.31 What is the net work done on the box? Wnet = What is the change in kinetic energy for the box? AKE = How fast is the box moving just before it hits the rear door? Vfinal= J J X J J m

The overworked Amazon delivery person is driving up a steep hill with an incline of 26° when a box they forgot to secure starts sliding toward the back of the truck. The 3.5 kg box starts from rest near the drivers seat and slides 2.1 m along the floor to the rear door. The coefficient of kinetic friction between the box and the floor is 0.36. What is the work done by the weight of the box? Wmg = 31.6 What is the work done by the Normal Force? WN = 0 What is the work done by the frictional force? W₁ = 23.31 What is the net work done on the box? Wnet = What is the change in kinetic energy for the box? AKE = How fast is the box moving just before it hits the rear door? Vfinal= J J X J J m

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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The text is a physics problem involving an Amazon delivery person driving a truck up a steep hill with an incline of 26°. A 3.5 kg box slides 2.1 m along the truck floor to the rear door due to the incline and movement. The coefficient of kinetic friction between the box and the truck floor is 0.36. Here's a breakdown of the calculations required:

1. **Work Done by the Weight of the Box (W<sub>mg</sub>)**:
- Calculated as 31.6 J (joules).

2. **Work Done by the Normal Force (W<sub>N</sub>)**:
- This value is 0 J, as the normal force does no work.

3. **Work Done by the Frictional Force (W<sub>fk</sub>)**:
- Calculated as 23.31 J. This is incorrect according to the text.

4. **Net Work Done on the Box (W<sub>net</sub>)**:
- The problem asks for this value to be calculated in joules.

5. **Change in Kinetic Energy of the Box (ΔKE)**:
- The problem asks for this value in joules.

6. **Final Speed of the Box (v<sub>final</sub>)**:
- The problem asks for this speed in meters per second (m/s).

The task involves applying concepts of work and energy, including the effects of friction, incline, and gravitational force on the box.

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