A mass A (.5kg)is placed on a smooth table with a string attached to it.The string goes over a frictions pulley and is connected to another mass of B (0.2kg). At t =0 the mass A is at a distance 2m from the end moving with a speed of 0.5m/s towards left,what will be its position and speed at t=1 sec?

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A mass A (.5kg)is placed on a smooth table with a string attached to it.The string goes over a frictions pulley and is connected to another mass of B (0.2kg). At t =0 the mass A is at a distance 2m from the end moving with a speed of 0.5m/s towards left,what will be its position and speed at t=1 sec?

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A mass A (.5kg)is placed on a smooth table with a string attached to i...

**Position and Speed of Mass A at t=1 sec**

To determine the position and speed of mass A at t=1 sec, we need to consider the system of masses and the forces acting on them.

**1. Free-Body Diagram:**

Let's start by drawing a free-body diagram for mass A:

- The weight of mass A acts vertically downwards (mg = 0.5 kg * 9.8 m/s^2 = 4.9 N).
- The tension in the string acts vertically upwards.
- There are no horizontal forces acting on mass A since the table is smooth.

**2. Forces Analysis:**

Since the string passes over a frictionless pulley and is connected to mass B, the tension in the string remains constant throughout the system. Therefore, the tension in the string pulling mass A is equal to the weight of mass B (T = 0.2 kg * 9.8 m/s^2 = 1.96 N).

**3. Newton's Second Law:**

Using Newton's second law, we can write the equation of motion for mass A:

ma = T - mg

Substituting the values, we get:

0.5 kg * a = 1.96 N - 4.9 N

0.5 kg * a = -2.94 N

a = -5.88 m/s^2

Since the acceleration is negative, it means that mass A is decelerating or moving towards the right.

**4. Calculating Position:**

To calculate the position of mass A at t=1 sec, we can use the kinematic equation:

x = x0 + v0t + (1/2)at^2

Given:
Initial position (x0) = 2 m
Initial velocity (v0) = -0.5 m/s
Acceleration (a) = -5.88 m/s^2
Time (t) = 1 sec

Substituting the values, we get:

x = 2 m + (-0.5 m/s)(1 sec) + (1/2)(-5.88 m/s^2)(1 sec)^2

x = 2 m - 0.5 m - 2.94 m

x = -1.44 m

Therefore, the position of mass A at t=1 sec is -1.44 m, indicating that it has moved 1.44 m to the left.

**5. Calculating Speed:**

To calculate the speed of mass A at t=1 sec, we can use the kinematic equation:

v = v0 + at

Given:
Initial velocity (v0) = -0.5 m/s
Acceleration (a) = -5.88 m/s^2
Time (t) = 1 sec

Substituting the values, we get:

v = -0.5 m/s + (-5.88 m/s^2)(1 sec)

v = -0.5 m/s - 5.88 m/s

v = -6.38 m/s

Therefore, the speed of mass A at t=1 sec is 6.38 m/s towards the left.

**Conclusion:**

At t=1 sec, the mass A will be at a position of -1.44 m (1.44 m to the left) and will have a speed of

Answered by Mohit Rajpoot · View profile

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A mass A (.5kg)is placed on a smooth table with a string attached to it.The string goes over a frictions pulley and is connected to another mass of B (0.2kg). At t =0 the mass A is at a distance 2m from the end moving with a speed of 0.5m/s towards left,what will be its position and speed at t=1 sec? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A mass A (.5kg)is placed on a smooth table with a string attached to it.The string goes over a frictions pulley and is connected to another mass of B (0.2kg). At t =0 the mass A is at a distance 2m from the end moving with a speed of 0.5m/s towards left,what will be its position and speed at t=1 sec? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mass A (.5kg)is placed on a smooth table with a string attached to it.The string goes over a frictions pulley and is connected to another mass of B (0.2kg). At t =0 the mass A is at a distance 2m from the end moving with a speed of 0.5m/s towards left,what will be its position and speed at t=1 sec?.

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