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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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Answer:

The acceleration of the object is 20 meters per second square = 20 m/s^2

Explanation:

Recall that acceleration is defined as the change in velocity divided the time it takes for the change. Therefore , if the object accelerates from rest (zero velocity) to 70 m/s , the change in velocity is (70 m/s - 0 m/s = 70 m/s)

which divided by the 3.5 seconds it took for the change, gives:

acceleration = (70 m/s  /  3.5 s ) = 20  m/s^2

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A softball is fouled off with a vertical velocity of 20 m/s and a horizontal velocity of 15 m/s. what is the resultant velocity
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Raindrops fall vertically at 7.5 m/s relative to the Earth. What does an observer in a car moving at 20.2 m/s in a straight line
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Answer:

vDP = 21.7454 m/s

θ = 200.3693°

Explanation:

Given

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vPE = 20.2 m/s

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Assume that

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vPE to be in direction of i

According to relative motion concept the velocity vDP is given by

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Substitute in (I) to get that

vDP = - 7.5 j - 20.2 i

The magnitude of vDP is given by

vDP = √((- 7.5)²+(- 20.2)²) m/s =  21.7454 m/s

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A- 1000 m/s2<br> Xi-0m<br> Xf-0.75m<br> Vf-?
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Answer:

The final velocity of the object is,  v_{f} = 27 m/s    

Explanation:

Given,

The acceleration of the object, a = 1000 m/s²

The initial displacement of the object, x_{i} = 0 m

The final displacement of the object,  x_{f} = 0.75 m

The initial velocity of the object will be, v_{i} = o m/s

The final velocity of the object, v_{f} = ?

The average velocity of the object,

                                    v = ( x_{f} - x_{i} )/ t

                                      = 0.75 / t

The acceleration is given by the relation

                                     a = v / t

                                   1000 m/s² = 0.75 / t²

                                            t² = 7.5 x 10⁻⁴

                                            t = 0.027 s

Using the I equation of motion,

                                  v_{f} = u + at

Substituting the values

                                   v_{f} = 0 + 1000 x 0.027

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Hence, the final velocity of the object is,  v_{f} = 27 m/s          

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