A shopper in a supermarket pushes a cart with a force of 35 N directed

Yolanda Jorge

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 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 50.0m length aisle.

Answer & Explanation

Elizabeth Witte

Elizabeth Witte

Beginner2021-11-23Added 24 answers

Given: 
Fapp=35N 
θ=25 Down the horizontal. 
x=50m 
We are aware that the work law is: W=Fcosθd=Fparalld 
In this case Fparall=Fappcos25° 
Wman=Fappcos25°x=35x×cos25°×50 
Wman=1.58kJ

Jennifer Hill

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=Fd
W=Fdcosθ
W=35N×50m×cos25°
W=1586.04J
Therefore, the work done by the shopper in moving the cart down the aisle is 1586.04 J.

star233

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·d·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=35N·50.0m·cos(25)
Calculating the cosine of 25 degrees and multiplying it by the other values, we get:
W=35N·50.0m·cos(25)1713.6J
Therefore, the work done by the shopper as she moves down the 50.0-m length aisle is approximately 1713.6 joules (J).
xleb123

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·d·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=35N·50.0m·cos(25)
Evaluating the cosine of 25 degrees and multiplying it by the other values, we find:
W35N·50.0m·cos(25)1713.6J
Hence, the work done by the shopper as she moves down the 50.0-meter length aisle is approximately 1713.6 joules (J).

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?