SOLUTION: a person pushes a 20 kg shopping car at a constant velocity for a distance of 33m. she pushes the handle in a direction 30 degrees to the horizontal. A 52N frictional force opposes
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Question 1170795: a person pushes a 20 kg shopping car at a constant velocity for a distance of 33m. she pushes the handle in a direction 30 degrees to the horizontal. A 52N frictional force opposes the motion of the cart.
how large is the force in the handle that pushes the cart forward (60.04N)
determine the work done by
a) the pushing force (1.72*10^3J)
b) the frictional force (-1.72*10^3J)
c) the gravational force (0J)
could u show me steps on solving!
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step by step.
**1. Finding the Pushing Force**
* **Forces Involved:**
* Applied force (F) at 30 degrees to the horizontal.
* Frictional force (f) = 52 N, opposing motion.
* Since the cart moves at a constant velocity, the net force in the horizontal direction is zero.
* **Horizontal Component of Applied Force:**
* F_x = F * cos(30°)
* **Equilibrium Condition:**
* F_x = f
* F * cos(30°) = 52 N
* **Solving for F:**
* F = 52 N / cos(30°)
* F = 52 N / (√3 / 2)
* F ≈ 60.04 N
**2. Work Done by the Pushing Force**
* **Work Done Formula:**
* W = F * d * cos(θ)
* Where:
* W is work done.
* F is the force.
* d is the distance.
* θ is the angle between the force and the displacement.
* **Calculation:**
* W = 60.04 N * 33 m * cos(30°)
* W ≈ 1716.9 J
* W ≈ 1.72 × 10³ J
**3. Work Done by the Frictional Force**
* **Frictional Force:**
* f = 52 N, opposing motion.
* **Work Done by Friction:**
* W_f = f * d * cos(180°)
* The angle between the frictional force and the displacement is 180 degrees because they are in opposite directions.
* W_f = 52 N * 33 m * (-1)
* W_f = -1716 J
* W_f ≈ -1.72 × 10³ J
**4. Work Done by the Gravitational Force**
* **Gravitational Force:**
* Fg = mg (where m is mass and g is acceleration due to gravity).
* **Work Done by Gravity:**
* W_g = Fg * d * cos(θ)
* The gravitational force acts vertically downwards, while the displacement is horizontal.
* Therefore, the angle between the gravitational force and the displacement is 90 degrees.
* W_g = Fg * d * cos(90°)
* W_g = Fg * d * 0
* W_g = 0 J
**Summary of Answers:**
* **Pushing Force:** 60.04 N
* **Work Done by Pushing Force:** 1.72 × 10³ J
* **Work Done by Frictional Force:** -1.72 × 10³ J
* **Work Done by Gravitational Force:** 0 J
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