**PRACTICE IT**Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of 1.30 × 10² N on the sled by pulling on the rope.(a) How much work does he do on the sled if the rope is horizontal to the ground (θ = 0° in the figure) and he pulls the sled 5.40 m?[Input box] J(b) How much work does he do on the sled if θ = 30.0° and he pulls the sled the same distance? (Treat the sled as a point particle, so details such as the point of attachment of the rope make no difference.)[Input box] J(c) At a coordinate position of 5.40 m, the Eskimo lets up on the applied force. A friction force of 50.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.40 m. How much work does friction do on the sled?[Input box] J---**EXERCISE**Suppose the Eskimo is pushing the same 50.0-kg sled across level terrain with a force of 55.0 N.(a) If he does 4.55 × 10² J of work on the sled while exerting the force horizontally, through what distance must he have pushed it?[Input box] m(b) If he exerts the same force at an angle of 50.0° with respect to the horizontal and moves the sled through the same distance, how much work does he do on the sled?[Input box] J**HINTS:** [GETTING STARTED] | [I'M STUCK!] **Example 5.1: Sledding Through the Yukon****Goal:** Apply the basic definitions of work done by a constant force.**Problem:** An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of \(1.20 \times 10^2 \, \text{N}\) on the sled by pulling on the rope. (a) How much work does he do on the sled if the rope is horizontal to the ground (\(\theta = 0^\circ\)) and he pulls the sled 5.00 m?(b) How much work does he do on the sled if \(\theta = 30.0^\circ\) and he pulls the sled the same distance? (Treat the sled as a point particle, so details such as the point of attachment of the rope make no difference.)(c) At a coordinate position of 12.4 m, the Eskimo lets up on the applied force. A friction force of 45.0 N between the ice and the sled brings the sled to rest at a coordinate position of 18.2 m. How much work does friction do on the sled?**Strategy:** Substitute the given values of \(F\) and \(\Delta x\) into the basic equations for work.---**Solution:****(A)** Find the work done when the force is horizontal.Use the proper equation, substituting the given values.\[ W = F_x \Delta x = (1.20 \times 10^2 \, \text{N})(5.00 \, \text{m}) = \boxed{6.00 \times 10^2 \, \text{J}} \]**(B)** Find the work done when the force is exerted at a \(30^\circ\) angle.Use the proper equation, again substituting the given values.\[ W = (F \cos \theta) d = (1.20 \times 10^2 \, \text{N})(\cos 30.0^\circ)(5.00 \, \text{m}) = \boxed{5.20 \times 10^2 \, \text{J}} \]**(C)** How much work does a friction force of \(45.

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Description: **PRACTICE IT**Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of 1.30 ×

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Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of 1.30 x 102 N on the sled by pulling on the rope. (a) How much work does he do on the sled if the rope is horizontal to the ground (0 = 0° in the figure) and he pulls the sled 5.40 m? J (b) How much work does he do on the sled if 0 = 30.0° and he pulls the sled the same distance? (Treat the sled as a point particle, so details such as the point of attachment of the rope make no difference.) J (c) At a coordinate position of 5.40 m, the Eskimo lets up on the applied force. A friction force of 50.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.40 m. How much work does friction do on the sled? J

Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of 1.30 x 102 N on the sled by pulling on the rope. (a) How much work does he do on the sled if the rope is horizontal to the ground (0 = 0° in the figure) and he pulls the sled 5.40 m? J (b) How much work does he do on the sled if 0 = 30.0° and he pulls the sled the same distance? (Treat the sled as a point particle, so details such as the point of attachment of the rope make no difference.) J (c) At a coordinate position of 5.40 m, the Eskimo lets up on the applied force. A friction force of 50.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.40 m. How much work does friction do on the sled? J

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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Topic Video

Question

**PRACTICE IT**

Use the worked example above to help you solve this problem. An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of 1.30 × 10² N on the sled by pulling on the rope.

(a) How much work does he do on the sled if the rope is horizontal to the ground (θ = 0° in the figure) and he pulls the sled 5.40 m?

[Input box] J

(b) How much work does he do on the sled if θ = 30.0° and he pulls the sled the same distance? (Treat the sled as a point particle, so details such as the point of attachment of the rope make no difference.)

[Input box] J

(c) At a coordinate position of 5.40 m, the Eskimo lets up on the applied force. A friction force of 50.0 N between the ice and the sled brings the sled to rest at a coordinate position of 10.40 m. How much work does friction do on the sled?

[Input box] J

---

**EXERCISE**

Suppose the Eskimo is pushing the same 50.0-kg sled across level terrain with a force of 55.0 N.

(a) If he does 4.55 × 10² J of work on the sled while exerting the force horizontally, through what distance must he have pushed it?

[Input box] m

(b) If he exerts the same force at an angle of 50.0° with respect to the horizontal and moves the sled through the same distance, how much work does he do on the sled?

[Input box] J

**HINTS:** [GETTING STARTED] | [I'M STUCK!]

**Example 5.1: Sledding Through the Yukon**

**Goal:**
Apply the basic definitions of work done by a constant force.

**Problem:**
An Eskimo returning from a successful fishing trip pulls a sled loaded with salmon. The total mass of the sled and salmon is 50.0 kg, and the Eskimo exerts a force of \(1.20 \times 10^2 \, \text{N}\) on the sled by pulling on the rope.

(a) How much work does he do on the sled if the rope is horizontal to the ground (\(\theta = 0^\circ\)) and he pulls the sled 5.00 m?

(b) How much work does he do on the sled if \(\theta = 30.0^\circ\) and he pulls the sled the same distance? (Treat the sled as a point particle, so details such as the point of attachment of the rope make no difference.)

(c) At a coordinate position of 12.4 m, the Eskimo lets up on the applied force. A friction force of 45.0 N between the ice and the sled brings the sled to rest at a coordinate position of 18.2 m. How much work does friction do on the sled?

**Strategy:**
Substitute the given values of \(F\) and \(\Delta x\) into the basic equations for work.

---

**Solution:**

**(A)** Find the work done when the force is horizontal.

Use the proper equation, substituting the given values.
\[ W = F_x \Delta x = (1.20 \times 10^2 \, \text{N})(5.00 \, \text{m}) = \boxed{6.00 \times 10^2 \, \text{J}} \]

**(B)** Find the work done when the force is exerted at a \(30^\circ\) angle.

Use the proper equation, again substituting the given values.
\[ W = (F \cos \theta) d = (1.20 \times 10^2 \, \text{N})(\cos 30.0^\circ)(5.00 \, \text{m}) = \boxed{5.20 \times 10^2 \, \text{J}} \]

**(C)** How much work does a friction force of \(45.

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