part (c) and part (d) **Problem 2:**A 32 kg block is pulled along the ground for a distance of 5.2 m by a 63 N force at an angle of 18 degrees with respect to the horizontal. The coefficient of kinetic friction is 0.25.*Diagram:* A block on a horizontal surface with an arrow indicating the direction of the applied force at an angle above the horizontal line.**Part (a):** Find the work done, in joules (J), by the normal force exerted by the ground.\[ W = \_\_\_ \, J \]*Calculator Interface:* - Functions: sin(), cos(), tan(), cotan(), asin(), acos(), atan(), acotan(), sinh(), cosh(), tanh(), coth()- Options for input: Degrees or Radians- Number pad with additional keys: π, e, backspace, delete, clear- Buttons: Submit, Hint, Feedback, I give up! **Question: Work Calculation****Part (a):** Find the work done, in Joules (J), by the normal force exerted by the ground.$ W = $ [ ] J**Calculator Functionality:** - **Trigonometric and Hyperbolic Functions:** - $ \sin() $, $ \cos() $, $ \tan() $ - $ \cotan() $, $ \asin() $, $ \acos() $ - $ \atan() $, $ \acotan() $, $ \sinh() $ - $ \cosh() $, $ \tanh() $, $ \coth() $- **Numeric and Arithmetic Operators:** - Numbers: 0-9 - Decimal operations: $ +, -, \times, \div $ - Other: $ \pi, (, ), E $- **Angle Mode:** - Options: Degrees, Radians - **Interface Options:** - Submit - Hint - Feedback - Clear---**Part (b):** Find the work done, in Joules (J), by the applied force.**Part (c):** Find the work done, in Joules (J), by the force of gravity.**Part (d):** Find the work done, in Joules (J), by the force of friction.

Answered: Part (b) Find the work done, in J, by…

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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part (c) and part (d)

**Problem 2:**

A 32 kg block is pulled along the ground for a distance of 5.2 m by a 63 N force at an angle of 18 degrees with respect to the horizontal. The coefficient of kinetic friction is 0.25.

*Diagram:* A block on a horizontal surface with an arrow indicating the direction of the applied force at an angle above the horizontal line.

**Part (a):** Find the work done, in joules (J), by the normal force exerted by the ground.

\[ W = \_\_\_ \, J \]

*Calculator Interface:*

- Functions: sin(), cos(), tan(), cotan(), asin(), acos(), atan(), acotan(), sinh(), cosh(), tanh(), coth()
- Options for input: Degrees or Radians
- Number pad with additional keys: π, e, backspace, delete, clear
- Buttons: Submit, Hint, Feedback, I give up!

**Question: Work Calculation**

**Part (a):** Find the work done, in Joules (J), by the normal force exerted by the ground.

$ W = $ [ ] J

**Calculator Functionality:**
- **Trigonometric and Hyperbolic Functions:**
- $ \sin() $, $ \cos() $, $ \tan() $
- $ \cotan() $, $ \asin() $, $ \acos() $
- $ \atan() $, $ \acotan() $, $ \sinh() $
- $ \cosh() $, $ \tanh() $, $ \coth() $

- **Numeric and Arithmetic Operators:**
- Numbers: 0-9
- Decimal operations: $ +, -, \times, \div $
- Other: $ \pi, (, ), E $

- **Angle Mode:**
- Options: Degrees, Radians

- **Interface Options:**
- Submit
- Hint
- Feedback
- Clear

---

**Part (b):** Find the work done, in Joules (J), by the applied force.

**Part (c):** Find the work done, in Joules (J), by the force of gravity.

**Part (d):** Find the work done, in Joules (J), by the force of friction.

Expert Solution

Given data:

The mass of the block, $m = 32 kg$

The horizontal displacement of the block, $d = 5.2 m$

The applied force, $F_{a} = 63 N$

The angle between the applied force and the horizontal, $θ = 18 °$

The coefficient of kinetic friction, $μ = 0.25$

Solution for (c):

The force of gravity is directed vertically downwards and the displacement of the block is in a horizontal direction, hence the angle between the direction of the force due to gravity and the displacement of the block is 270° (in a counterclockwise direction).

The work done by a force is a dot product of the force and the displacement. Therefore,

$\begin{array}{rcl} W_{g} & = & {\vec{F}}_{g} ∙ \vec{d} \\ W_{g} & = & F_{g} d \cos (270 °) \\ W_{g} & = & 0 - - - (∵ \cos (270 °) = 0) \end{array}$

Therefore, the work done by the force of gravity on the block is 0 J.

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