Anuja is holding a stuffed dog, with a mass of 0.30 kg, when Derek decides that he wants it and tries to pull it away from Abuja
1 answer:
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Answer:
2m₁m₃g / (m₁ + m₂ + m₃)
Explanation:
I assume the figure is the one included in my answer.
Draw a free body diagram for each mass.
m₁ has a force T₁ up and m₁g down.
m₂ has a force T₁ up, T₂ down, and m₂g down.
m₃ has a force T₂ up and m₃g down.
Assume that m₁ accelerates up and m₂ and m₃ accelerate down.
Sum of the forces on m₁:
∑F = ma
T₁ − m₁g = m₁a
T₁ = m₁g + m₁a
Sum of the forces on m₂:
∑F = ma
T₁ − T₂ − m₂g = m₂(-a)
T₁ − T₂ − m₂g = -m₂a
(m₁g + m₁a) − T₂ − m₂g = -m₂a
m₁g + m₁a + m₂a − m₂g = T₂
(m₁ − m₂)g + (m₁ + m₂)a = T₂
Sum of the forces on m₃:
∑F = ma
T₂ − m₃g = m₃(-a)
T₂ − m₃g = -m₃a
a = g − (T₂ / m₃)
Substitute:
(m₁ − m₂)g + (m₁ + m₂) (g − (T₂ / m₃)) = T₂
(m₁ − m₂)g + (m₁ + m₂)g − ((m₁ + m₂) / m₃) T₂ = T₂
(m₁ − m₂)g + (m₁ + m₂)g = ((m₁ + m₂ + m₃) / m₃) T₂
m₁g − m₂g + m₁g + m₂g = ((m₁ + m₂ + m₃) / m₃) T₂
2m₁g = ((m₁ + m₂ + m₃) / m₃) T₂
T₂ = 2m₁m₃g / (m₁ + m₂ + m₃)
Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :
Work done on the wagon is given by the product of net force and displacement. It is given by :
W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.