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igor_vitrenko [27]
3 years ago
10

A cart of mass M is attached to an ideal spring that can stretch and compress equally well. The cart and spring rest on a smooth

horizontal track. The cart is pulled to position A and released. The cart then moves toward position E, where it reverses direction and returns again to position A.
a) show with the sketch of a position diagram how the velocity of the cart changes between the points indicated
b) The dots below represent the cart at different labeled positions. Draw free body diagrams showing and labeling all the forces exerted on the cart at each labeled position. Draw the relative lengths of all vectors to reflect the relative magnitude of all forces. You will create one set of sketches for when the cart is moving to the right and a second set for when the car is moving to the left.
c) do the diagrams above indicate whether the cart is moving left or right? Justify your claim with evidence.
d) at each position, compare the direction of the net force exerted by the spring on the cart and the carts displacement from equilibrium when at that position. Note that this question is not asking whether the cart is moving right or left. Use these results to briefly explain why your claim in part C makes sense.

Physics
1 answer:
Alexxx [7]3 years ago
4 0

We have to remember a point , which is ' the cart or spring rest on a smooth horizontal track ' , i.e., any frictional force doesn't take place.

<u>Explanation:</u>

a) According to the question the cart is pulled to position A and released, i.e., the velocity of the cart at A initially (say time,t=0) is 0 m/s ,then moves toward position E, where it reverses direction and returns again to position A , in the 2nd phase cart moves along A to E , the cart's velocity increase and again goes to zero at point E and again change the direction, hence

( File has been attached)

b) Let's , the distance between two consecutive points is x meter and the spring constant is k N.m  

c) ( File has been attached)

d) Movinf Right

e) Moving left

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<u>Hello and Good Morning/Afternoon</u>:

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<u>To balance this equation</u>:

⇒ must ensure that there is an equal number of elements on both sides of the equation at all times

<u>Let's start balancing:</u>

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<u>Let's check and make sure we got the answer:</u>

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<em>                 2 Carbon                ⇔                    2 Carbon</em>

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