Get the solution to "A consumer in a grocery store pushes a..."

High School
Geometry
Trigonometry
Right triangles and trigonometry

Yolanda Jorge

Answered question

2021-11-22

A consumer in a grocery store pushes a cart with a force of 35 N directed at an angle of $25^{\circ}$ below the horizontal. The force is just enough to overcome various frictional forces, so the cart moves at a steady pace. Find the work done by the shopper as she moves down a $50.0 - m$ length aisle.

Answer & Explanation

Elizabeth Witte

Beginner2021-11-23Added 24 answers

Given:
$F_{a p p} = 35 N$
$θ = 25^{\circ}$ Down the horizontal.
$△ x = 50 m$
We are aware that the work law is: $W = F \cos θ d = F_{p a r a l \leq l} d$
In this case $F_{p a r a l \leq l} = F_{a p p} \cos 25 °$
$W_{m a n} = F_{a p p} \cos 25 ° △ x = 35 x \times \cos 25 ° \times 50$
$W_{m a n} = 1.58 k J$

Jennifer Hill

Beginner2021-11-24Added 10 answers

The expression for the work done by the shopper in moving the cart down the aisle is given by,
$W = F \cdot d$
$W = F d \cos θ$
$W = 35 N \times 50 m \times \cos 25 °$
$W = 1586.04 J$
Therefore, the work done by the shopper in moving the cart down the aisle is 1586.04 J.

star233

Skilled2023-05-10Added 403 answers

To solve this problem, we need to find the work done by the shopper as she moves down the aisle. The work done is given by the equation:

W = F \cdot d \cdot \cos (θ)

where:
-

W

represents the work done (in joules, J)
-

F

is the applied force (in newtons, N)
-

d

is the displacement (in meters, m)
-

θ

is the angle between the force and displacement vectors (in degrees)
In this case, the applied force is 35 N, the displacement is 50.0 m, and the angle

θ

is 25 degrees below the horizontal.
Substituting these values into the equation, we have:

W = 35 N \cdot 50.0 m \cdot \cos (25^{\circ})

Calculating the cosine of 25 degrees and multiplying it by the other values, we get:

W = 35 N \cdot 50.0 m \cdot \cos (25^{\circ}) \approx 1713.6 J

Therefore, the work done by the shopper as she moves down the 50.0-m length aisle is approximately 1713.6 joules (J).

xleb123

Skilled2023-05-10Added 181 answers

Answer:
1713.6 joules (J)
Explanation:
To determine the work done by the shopper as she moves down the aisle, we can use the equation:

W = F \cdot d \cdot \cos (θ)

Given that the applied force is 35 N, the displacement is 50.0 m, and the angle

θ

is 25 degrees below the horizontal, we can substitute these values into the equation:

W = 35 N \cdot 50.0 m \cdot \cos (25^{\circ})

Evaluating the cosine of 25 degrees and multiplying it by the other values, we find:

W \approx 35 N \cdot 50.0 m \cdot \cos (25^{\circ}) \approx 1713.6 J

Hence, the work done by the shopper as she moves down the 50.0-meter length aisle is approximately 1713.6 joules (J).

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