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Alex73 [517]
3 years ago
8

A boy is pulling a toy wagon through a neighborhood at a speed of 0.75 m/s. A caterpillar in the wagon is crawling toward the re

ar of the wagon at a rate of 2.0 cm/s. What is the caterpillar’s velocity relative to the ground?
Physics
1 answer:
Zepler [3.9K]3 years ago
4 0

Answer:

0.73 m/s in the forward direction

Explanation:

The caterpillar's velocity relative to the ground is given by:

v' = v_w + v

where

v_w is the velocity of the wagon with respect to the ground

v is the velocity of the caterpillar in the wagon's reference frame

In this situation we have

v_w = +0.75 m/s

v=-2.0 cm/s = -0.02 m/s where the negative sign is due to the fact that the caterpillar is moving backward

Substituting,

v'=+0.75-0.02=+0.73 m/s

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Leaning over and Pushing the heavy box from the floor, the push will be divided in to two parts, one is horizontal that can help the box move, and one is vertically downwards, which increases the downward force of the heavy object (an addition to the gravity) and thus increases friction, making it very hard to push.  When you push at certain angle, you are exhibiting two forces as shown in diagram 1.

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Case 2: <u>Pulling</u> Diagram 2

Pulling on a rope similar object at the same angle, the pull can be divided into two parts, one is horizontal that can help the box move, and one is vertically upwards, which decreases the downwards force of the box (a subtraction in the gravity) and thus decreases friction, making it very easy to pull. When you pull at a certain angle, you are exhibiting two forces as shown in diagram 2.

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So, in the case of pushing, it adds an extra weight on the object, which results in difficulty to push that object at the same angle.  In case of pulling, the upward perpendicular force, it tries to lift the  object upward and divided the weight partially. Thus making it easier to move the object at same angle.

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Explanation:

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y(t) = h + Vo sin θt - 1/2gt²

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