A mass m1 = 5.0 kg rests on a frictionless table and connected by a massless string to another mass m2 = 5.9 kg. A force of magnitude F = 40.0 N pulls m1 to the left a distance d = 0.77 m.
1)How much work is done by the force F on the two block system?
2)What is the final speed of the two blocks?
3)How much work is done by the tension (in-between the blocks) on block m2?
4)What is the tension in the string
To find the answers to these questions, we need to use the concepts of work, potential energy, and conservation of energy. Here's how you can solve each question:
1) How much work is done by the force F on the two block system?
The work done by a force is given by the equation W = F * d * cos(theta), where F is the magnitude of the force, d is the displacement, and theta is the angle between the force and displacement vectors. In this case, the force F is pulling m1 to the left, so theta is 180 degrees (or pi radians).
Given:
F = 40.0 N
d = 0.77 m
theta = 180 degrees = pi radians
Using the formula, we have:
W = 40.0 N * 0.77 m * cos(pi) = -30.8 J
Therefore, the work done by the force F on the two-block system is -30.8 J.
2) What is the final speed of the two blocks?
To find the final speed, we need to use the principle of conservation of energy. The work done by the external force is equal to the change in kinetic energy of the system.
Initially, both blocks are at rest, so their initial kinetic energy is zero. Therefore, the work done by the force F is equal to the final kinetic energy of the system.
Using the work-energy theorem, we have:
W = (1/2) * (m1 + m2) * v^2
Given:
W = -30.8 J
m1 = 5.0 kg
m2 = 5.9 kg
Since the two blocks move together, their final velocity (v) is the same.
Rearranging the equation, we have:
v^2 = -2 * (W / (m1 + m2))
Substituting the values, we get:
v^2 = -2 * (-30.8 J / (5.0 kg + 5.9 kg))
v^2 = 9.828 J / 10.9 kg
v^2 = 0.9002 J/kg
v = sqrt(0.9002) m/s
Therefore, the final speed of the two blocks is approximately 0.948 m/s.
3) How much work is done by the tension (in-between the blocks) on block m2?
In this case, the tension force acts in the opposite direction of the displacement of m2. Thus, the work done by the tension force is negative.
Using the same formula as in question 1 (W = F * d * cos(theta)), and noticing that the distance d is the same for both blocks, we have:
W = -F * d * cos(theta)
Given:
F = tension force
d = 0.77 m
theta = 180 degrees = pi radians
Therefore, the work done by the tension force on block m2 is -40.0 N * 0.77 m * cos(pi) = 30.8 J.
4) What is the tension in the string?
Since the two blocks are connected by a massless string, the tension force in the string is the same throughout. This means the tension force is equal to the force applied to m1, which is 40.0 N.
Therefore, the tension in the string is 40.0 N.