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Allisa [31]
3 years ago
9

A crate is being lifted into a truck. If it is moved with a 2470 N force and 4850 J of work is done, then how far is the crate b

eing lifted?
Physics
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:1.96m

Explanation:

Work=4850J

Force=2470N

Distance =work ➗ force

Distance =4850 ➗ 2470

Distance =1.96m

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Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
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 K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

          p₀ = 0

final moment

          p_f = m_a v_a + m_b v_b

          p₀ = p_f

          0 = m_a v_a + m_b v_b

tells us that

          m_a = 8 m_b

         

           0 = 8 m_b v_a + m_b v_b

           v_b = - 8 v_a                    (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

           Em₀ = K₀ = 73 J

final point. After separating the body

          Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

           K₀ = K_f

           73 = ½ m_a (v_a² + v_b² / 8)

           

we substitute equation 1

           73 = ½ m_a (v_a² + 8² v_a² / 8)

           73 = ½ m_a (9 v_a²)

           73/9 = ½ m_a (v_a²) = K_a

            K_a = 8,111 J

3 0
3 years ago
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