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A mass of 4.5kg is pulled along a level track by a 1.5kg mass that hangs vertically. The masses are attached by a light smooth pulley. calculate the acceleration of 4.5 kg mass and the distance traveled by it in 2 seconds , given that the coefficient of friction between the 4.5 kg mass and the track is 0.2?
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A mass of 4.5kg is pulled along a level track by a 1.5kg mass that ha...
**Given:**
- Mass of the 4.5 kg mass (m1) = 4.5 kg
- Mass of the hanging mass (m2) = 1.5 kg
- Coefficient of friction between the 4.5 kg mass and the track (μ) = 0.2
**To find:**
- Acceleration of the 4.5 kg mass (a)
- Distance traveled by the 4.5 kg mass in 2 seconds (s)
**Assumptions:**
- The pulley is light and smooth, so there is no friction or mass associated with it.
- The hanging mass is negligible compared to the 4.5 kg mass, so the tension in the string is the same on both sides.
- The system is isolated, i.e., there are no external forces acting on it.
**Solution:**
1. **Calculating the tension in the string:**
- Since the hanging mass is in equilibrium, the tension in the string (T) is equal to m2 * g, where g is the acceleration due to gravity (9.8 m/s^2).
- T = m2 * g = 1.5 kg * 9.8 m/s^2 = 14.7 N
2. **Calculating the force of friction:**
- The force of friction between the 4.5 kg mass and the track (Ff) can be calculated using the equation Ff = μ * N, where N is the normal force.
- The normal force (N) is equal to the weight of the 4.5 kg mass, which is N = m1 * g = 4.5 kg * 9.8 m/s^2 = 44.1 N
- Ff = μ * N = 0.2 * 44.1 N = 8.82 N
3. **Calculating the net force on the 4.5 kg mass:**
- The net force (Fnet) acting on the 4.5 kg mass is the difference between the tension in the string (T) and the force of friction (Ff).
- Fnet = T - Ff = 14.7 N - 8.82 N = 5.88 N
4. **Calculating the acceleration of the 4.5 kg mass:**
- The acceleration (a) of the 4.5 kg mass can be calculated using Newton's second law, Fnet = m1 * a.
- 5.88 N = 4.5 kg * a
- a = 5.88 N / 4.5 kg ≈ 1.31 m/s^2
5. **Calculating the distance traveled by the 4.5 kg mass in 2 seconds:**
- The distance traveled (s) can be calculated using the equation s = ut + 0.5 * a * t^2, where u is the initial velocity (assumed to be 0 m/s) and t is the time.
- s = 0 + 0.5 * 1.31 m/s^2 * (2 s)^2 = 0.5 * 1.31 m/s^2 * 4 s^2 = 2.62 m
**Answer
- Mass of the 4.5 kg mass (m1) = 4.5 kg
- Mass of the hanging mass (m2) = 1.5 kg
- Coefficient of friction between the 4.5 kg mass and the track (μ) = 0.2
**To find:**
- Acceleration of the 4.5 kg mass (a)
- Distance traveled by the 4.5 kg mass in 2 seconds (s)
**Assumptions:**
- The pulley is light and smooth, so there is no friction or mass associated with it.
- The hanging mass is negligible compared to the 4.5 kg mass, so the tension in the string is the same on both sides.
- The system is isolated, i.e., there are no external forces acting on it.
**Solution:**
1. **Calculating the tension in the string:**
- Since the hanging mass is in equilibrium, the tension in the string (T) is equal to m2 * g, where g is the acceleration due to gravity (9.8 m/s^2).
- T = m2 * g = 1.5 kg * 9.8 m/s^2 = 14.7 N
2. **Calculating the force of friction:**
- The force of friction between the 4.5 kg mass and the track (Ff) can be calculated using the equation Ff = μ * N, where N is the normal force.
- The normal force (N) is equal to the weight of the 4.5 kg mass, which is N = m1 * g = 4.5 kg * 9.8 m/s^2 = 44.1 N
- Ff = μ * N = 0.2 * 44.1 N = 8.82 N
3. **Calculating the net force on the 4.5 kg mass:**
- The net force (Fnet) acting on the 4.5 kg mass is the difference between the tension in the string (T) and the force of friction (Ff).
- Fnet = T - Ff = 14.7 N - 8.82 N = 5.88 N
4. **Calculating the acceleration of the 4.5 kg mass:**
- The acceleration (a) of the 4.5 kg mass can be calculated using Newton's second law, Fnet = m1 * a.
- 5.88 N = 4.5 kg * a
- a = 5.88 N / 4.5 kg ≈ 1.31 m/s^2
5. **Calculating the distance traveled by the 4.5 kg mass in 2 seconds:**
- The distance traveled (s) can be calculated using the equation s = ut + 0.5 * a * t^2, where u is the initial velocity (assumed to be 0 m/s) and t is the time.
- s = 0 + 0.5 * 1.31 m/s^2 * (2 s)^2 = 0.5 * 1.31 m/s^2 * 4 s^2 = 2.62 m
**Answer
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A mass of 4.5kg is pulled along a level track by a 1.5kg mass that hangs vertically. The masses are attached by a light smooth pulley. calculate the acceleration of 4.5 kg mass and the distance traveled by it in 2 seconds , given that the coefficient of friction between the 4.5 kg mass and the track is 0.2?
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A mass of 4.5kg is pulled along a level track by a 1.5kg mass that hangs vertically. The masses are attached by a light smooth pulley. calculate the acceleration of 4.5 kg mass and the distance traveled by it in 2 seconds , given that the coefficient of friction between the 4.5 kg mass and the track is 0.2? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A mass of 4.5kg is pulled along a level track by a 1.5kg mass that hangs vertically. The masses are attached by a light smooth pulley. calculate the acceleration of 4.5 kg mass and the distance traveled by it in 2 seconds , given that the coefficient of friction between the 4.5 kg mass and the track is 0.2? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mass of 4.5kg is pulled along a level track by a 1.5kg mass that hangs vertically. The masses are attached by a light smooth pulley. calculate the acceleration of 4.5 kg mass and the distance traveled by it in 2 seconds , given that the coefficient of friction between the 4.5 kg mass and the track is 0.2?.
A mass of 4.5kg is pulled along a level track by a 1.5kg mass that hangs vertically. The masses are attached by a light smooth pulley. calculate the acceleration of 4.5 kg mass and the distance traveled by it in 2 seconds , given that the coefficient of friction between the 4.5 kg mass and the track is 0.2? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A mass of 4.5kg is pulled along a level track by a 1.5kg mass that hangs vertically. The masses are attached by a light smooth pulley. calculate the acceleration of 4.5 kg mass and the distance traveled by it in 2 seconds , given that the coefficient of friction between the 4.5 kg mass and the track is 0.2? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mass of 4.5kg is pulled along a level track by a 1.5kg mass that hangs vertically. The masses are attached by a light smooth pulley. calculate the acceleration of 4.5 kg mass and the distance traveled by it in 2 seconds , given that the coefficient of friction between the 4.5 kg mass and the track is 0.2?.
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