Answer:
Value of magnitude of acceleration will be
Explanation:
It is given that when a stone is dropped it takes 2.2 sec to reach the ground
(a) As the stone is dropped from the top of building
So its initial velocity will be 0 m /sec
(b) As the stone is free falling and there is no external force applied on it so its acceleration will be equal to acceleration due to gravity
So value of magnitude of acceleration will be equal to
Answer:
(a)Look at the attached graphic
(b)
(b)-1 Equation 1 : m1= 5kg
50-F1= 5 *a
(b)-2 Equation 2 : m2= 3kg
F1-F2= 3 *a
(b)-3 Equation 3 : m3= 2kg
F2 = 2*a
(c) F1 =25 N
(d) F2 =10 N
Explanation:
We apply Newton's second law:
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
(a) Draw the free-body diagrams for each of the boxes
Look at the attached graphic
(b) Write Newton’s equation for each mass along the horizontal direction.
Data: m1= 5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg
<em>Look</em> <em>m1 free-body diagram:</em>
∑Fx = m1*a
50-F1= 5 *a Equation 1
<em>Look</em> <em>m2 free-body diagram:</em>
∑Fx = m2*a
F1-F2= 3 *a Equation 2
<em>Look</em> <em>m3 free-body diagram:</em>
∑Fx = m3*a
F2 = 2*a Equation 3
(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?
<em>Look</em> <em>Free body diagram of the mass set</em>
∑Fx = m*a m= m1+m2+m3= 5+3+2 = 10 kg
50 = 10*a
a= 50/10 = 5 m/s²
We replace a = 5 m/s² in the equation 1:
50-F1= 5 *5
50-25= F1
F1 = 25 N
<em> (d) </em><em>What magnitude force does the 3.0-kg box exert on the 2.0kg box?</em>
We replace a= 5 m/s² in the equation 3
F2 = 2*5 = 10 N
