Question

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 1.00×10² 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 4.90 m?
(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.)
(c) at a coordinate position of 4.90 m, the eskimo lets up on the applied force. a friction force of 41.0 n between the ice and the sled brings the sled to rest at a coordinate position of 9.90 m. how much work does friction do on the sled?
exercise
hints: getting started | im stuck!
suppose the eskimo is pushing the same 50.0 - kg sled across level terrain with a force of 45.0 n.
(a) if he does 4.15×10² j of work on the sled while exerting the force horizontally, through what distance must he have pushed it?
(b) if he exerts the same force at an angle of 30.0° with respect to the horizontal and moves the sled through the same distance, how much work does he do on the sled?

Explanation:

Step1: Recall the work - formula

The work - done formula is $W = F\cdot d\cdot\cos\theta$, where $W$ is work, $F$ is the force applied, $d$ is the displacement, and $\theta$ is the angle between the force and the displacement.

Part (a)

Step1: Identify the values

Given $F = 1.00\times10^{2}\ N$, $d = 4.90\ m$, and $\theta = 0^{\circ}$, and $\cos(0^{\circ})=1$.

Step2: Calculate the work

$W = F\cdot d\cdot\cos\theta=(1.00\times 10^{2}\ N)\times(4.90\ m)\times1 = 490\ J$

Part (b)

Step1: Identify the values

Given $F = 1.00\times10^{2}\ N$, $d = 4.90\ m$, and $\theta = 30.0^{\circ}$, and $\cos(30^{\circ})=\frac{\sqrt{3}}{2}\approx0.866$.

Step2: Calculate the work

$W = F\cdot d\cdot\cos\theta=(1.00\times 10^{2}\ N)\times(4.90\ m)\times0.866 = 424.34\ J$

Part (c)

Step1: Identify the values

The friction force $F_f=41.0\ N$, the displacement due to the friction force $d_f=(9.90 - 4.90)\ m = 5.00\ m$, and the angle between the friction force and the displacement $\theta = 180^{\circ}$ (since the friction force opposes the motion), and $\cos(180^{\circ})=- 1$.

Step2: Calculate the work

$W = F_f\cdot d_f\cdot\cos\theta=(41.0\ N)\times(5.00\ m)\times(-1)=-205\ J$

Exercise (a)

Step1: Rearrange the work - formula

From $W = F\cdot d\cdot\cos\theta$, when $\theta = 0^{\circ}$ and $\cos(0^{\circ}) = 1$, we can solve for $d$: $d=\frac{W}{F}$.

Step2: Identify the values

Given $W = 4.15\times10^{2}\ J$ and $F = 45.0\ N$.

Step3: Calculate the distance

$d=\frac{4.15\times 10^{2}\ J}{45.0\ N}\approx9.22\ m$

Exercise (b)

Step1: Identify the values

Given $F = 45.0\ N$, $d\approx9.22\ m$ (from part (a)), and $\theta = 30.0^{\circ}$, $\cos(30^{\circ})\approx0.866$.

Step2: Calculate the work

$W = F\cdot d\cdot\cos\theta=(45.0\ N)\times(9.22\ m)\times0.866 = 360.99\ J$

Answer:

(a) $490\ J$
(b) $424.34\ J$
(c) $-205\ J$
Exercise (a) $9.22\ m$
Exercise (b) $360.99\ J$