Discover the answers you need at Westonci.ca, a dynamic Q&A platform where knowledge is shared freely by a community of experts. Our platform provides a seamless experience for finding reliable answers from a knowledgeable network of professionals. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
Answer:
v = 6.195 m / s
Explanation:
For this exercise we can use the conservation of energy, for the system formed by the ball and the pulley
starting point. Higher before releasing the system
Em₀ = U = M g h
final point. When the ball has lowered h = 2
Em_f = K = ½ M v² + ½ I w²
the energy is preserved
Em₀ = Em_f
M g h = ½ M v² + ½ I w²
angular and linear velocity are related
v = w r
w = v / r
indicate that the moment of inertia is
I = ½ m r²
we substitute
M g h = ½ M v² + ½ (½ m r²) (v/r) ²
½ v² (M + [tex]\frac{1}{2}[/tex] m) = M g h
v² = [tex]2gh \ \frac{M}{M + \frac{m}{2} }[/tex]
let's calculate
v = [tex]\sqrt{ 2 \ 9.8 \ 2 \ \frac{70}{70 + 1.5} }[/tex]
v = 6.195 m / s
Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
