Answer:
A. 4,9 m/s2
B. 2,0 m/s2
C. 120 N
Explanation:
In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:
[tex]\sum F_y=m_1*a_m_i_n = T-m_1*g[/tex]
If the package is barely lifted, that means that T=m_2*g; then:
[tex]\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g[/tex]
Solving the equation for a_mín, we have:
[tex]a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2[/tex]
Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:
For the monkey: [tex]\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a[/tex]
For the package: [tex]\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a[/tex]
The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:
For the package: [tex]\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a[/tex]
We have two unknowns and two equations, so we can proceed. We can match both tensions and have:
[tex]m_1*a+m_1*g=m_2*g-m_2*a[/tex]
Solving a, we have
[tex](m_1+m_2)*a =(m_2 - m1)*g\\\\a=((m_2-m_1)/(m_1+m_2))*g \rightarrow a=((15kg-10kg)/(10kg+15kg))*9,8 m/s^2\\\\a= 2,0 m/s^2[/tex]
We can then replace this value of a in one for the sums of force and find the tension T:
[tex]T = m_1*a+m_1*g \rightarrow T=m_1*(a+g)\\\\T = 10kg*(2,0 m/s^2+9,8 m/s^2) \\\\T = 120 N[/tex]
