Two boxes m1 and m2 are connected by a lightweight cord and sitting on a frictionless surface. A horizontal force of 66 N pulls
on box m1 and then m2 is pulled along. If the acceleration of both boxes is 2.2 m/s2 and the mass of m1 is 12 kg, find the mass of m2
1 answer:
Answer:
m₂ = 18 kg
Explanation:
Conceptual analysis
We apply Newton's second law on masses m₁ and m₂
∑F = m × a Formula (1)
∑F: algebraic sum of forces in Newton (N)
m: mass in kg
a: acceleration in m/s²
T: cord tension in Newton (N)
Known data
m₁ = 12 kg
F = 66 N
a₁ = a₂ = 2.2 m/s²
Problem development
We can see in the attached figure the free body diagram of the masses and we replace the data in formula (1), considering that positive forces are the ones that go in the same direction of acceleration::
Mass 1 (m₁) free body diagram:
∑Fₓ = m₁ × a₁
66 - T = 12 × 2.2
66 - T = 26.4
66 - 26.4 = T
T = 39.6 N
Mass 2 (m₂) free body diagram:
∑Fₓ = m₂ × a₂
T = m₂ × 2.2
39.6 = m₂ × 2.2
m₂ = 39.6 ÷ 2.2
m₂ = 18 kg
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Answer:
The x component of the resultant force is -7.27N.
Explanation:
To obtain the x component of the resultant force, first we have to know the x components of the other forces. To do this, we just have to do some trigonometry:
Since both vectors are in the left side of the y-axis, they have a negative x component. So:
Finally, we sum both components to obtain the component of the resultant force:
In words, the x component of the resultant force is -7.27N.
The electric force acting on the charge is given by the charge multiplied by the electric field intensity:
where in our problem
and 
, so the force is
The initial kinetic energy of the particle is zero (because it is at rest), so its final kinetic energy corresponds to the work done by the electric force for a distance of x=4 m:
where in our problem
The initial kinetic energy of the particle is zero (because it is at rest), so its final kinetic energy corresponds to the work done by the electric force for a distance of x=4 m:
Answer:
The solution and the explanation are in the Explanation section.
Explanation:
According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:
TA = W * (a * cosθ)
The torque due to effort at C point is:
TC = E * (b * cosθ)
The net torque is equal to 0, we have:
Tnet = 0
W * (a * cosθ) - E * (b * cosθ) = 0
From the figure, you can observe that a/b < 1, thus E < W
