A mass m1 = 4.1 kg rests on a frictionless table and connected by a massless string to another mass m2 = 5.9 kg. A force of magnitude F = 32.0 N pulls m1 to the left a distance d = 0.83 m.
1) How much work is done by the force F on the two block system?
2) How much work is done by the normal force on m1 and m2? = 0
3) What is the final speed of the two blocks?
4) How much work is done by the tension (in-between the blocks) on block m2?
5) What is the tension in the string?
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To answer these questions, we can use the principles of work and energy. Here are the step-by-step explanations to find the answers to each question:
1) To find the work done by the force F on the two-block system, we can use the formula:
Work = Force * Distance * cos(theta)
In this case, the force F is pulling the block m1 to the left, so the angle between the force and the direction of motion is 180 degrees (cos(180) = -1). Thus, the equation becomes:
Work = 32.0 N * 0.83 m * -1 = -26.56 J
So, the work done by the force F on the two-block system is -26.56 Joules.
2) The normal force is perpendicular to the direction of motion. Therefore, the work done by the normal force on both m1 and m2 is zero since no displacement occurs in the direction of the normal force.
3) To find the final speed of the two blocks, we can use the principle of conservation of mechanical energy. The initial mechanical energy of the system is equal to the final mechanical energy.
The initial mechanical energy is given by:
Initial Mechanical Energy = Kinetic Energy (m1) + Kinetic Energy (m2)
Since the blocks are initially at rest, the initial kinetic energy is zero. Therefore, the initial mechanical energy is zero.
The final mechanical energy is given by:
Final Mechanical Energy = Kinetic Energy (m1) + Kinetic Energy (m2)
We can calculate the kinetic energy of each block using the formula:
Kinetic Energy = (1/2) * mass * velocity^2
Since the two blocks are connected by a string, they move together and have the same final velocity.
So, we have:
Final Mechanical Energy = (1/2) * m1 * velocity^2 + (1/2) * m2 * velocity^2
Since the initial mechanical energy is zero, we can solve for the final velocity:
(1/2) * m1 * velocity^2 + (1/2) * m2 * velocity^2 = 0
(1/2) * (m1 + m2) * velocity^2 = 0
Therefore, the final velocity of the two blocks is 0 m/s.
4) The work done by the tension force on block m2 can be found using the same formula as in question 1:
Work = Force * Distance * cos(theta)
In this case, the force is the tension between the blocks, and there is no displacement in the direction of the tension force. Therefore, the work done by the tension force on block m2 is zero.
5) To find the tension in the string, we can use Newton's second law and equate the sum of the forces acting on block m1 to its mass times acceleration.
The forces acting on m1 are the tension force (T) to the right and the force F to the left. Therefore:
Sum of Forces = T - F = m1 * a
Since the blocks are connected and move together, the acceleration of both blocks will be the same. Therefore:
m1 * a = (m1 + m2) * a
Simplifying the equation:
T - 32.0 N = (4.1 kg + 5.9 kg) * a
T - 32.0 N = 10 kg * a
Since the blocks are at rest (as found in question 3), the acceleration is zero. Therefore, the tension force can be calculated as:
T - 32.0 N = 0
T = 32.0 N
So, the tension in the string is 32.0 Newtons.