Complete Question
The diagram of this question is shown on the first uploaded image
Answer:
The distance the block slides before stopping is [tex]d = 0.313 \ m[/tex]
Explanation:
The free body diagram for the diagram in the question is shown
From the diagram the angle is [tex]\theta = 25 ^o[/tex]
[tex]sin \theta = \frac{h}{d}[/tex]
Where [tex]h = h_b - h_a[/tex]
So [tex]d sin \theta = h_b - h_a[/tex]
From the question we are told that
The mass of the block is [tex]m = 20 \ kg[/tex]
The mass of the pendulum is [tex]m_p = 2 \ kg[/tex]
The velocity of the pendulum at the bottom of swing is [tex]v_p = 15 m/s[/tex]
The coefficient of restitution is [tex]e =0.7[/tex]
The coefficient of kinetic friction is [tex]\mu _k = 0.5[/tex]
The velocity of the block after the impact is mathematically represented as
[tex]v_2 f = \frac{m_b - em_p}{m_b + m_p} * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p[/tex]
Where [tex]v_2 i[/tex] is the velocity of the block before collision which is 0
[tex]= \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20} * 15[/tex]
Substituting value
[tex]v_2 f = 2.310\ m/s[/tex]
According to conservation of energy principle
The energy at point a = energy at point b
So [tex]PE_A + KE _A = PE_B + KE_B + E_F[/tex]
Where
[tex]PE_A[/tex] is the potential energy at A which is mathematically represented as
[tex]PE_A = m_b gh_a[/tex] = 0 at the bottom
[tex]KE _A[/tex] is the kinetic energy at A which is mathematically represented as
[tex]K_A = \frac{1}{2} m_b * v_2f^2[/tex]
[tex]PE_B[/tex] is the potential energy at B which is mathematically represented as
[tex]PE_B = m_b gh[/tex]
From the diagram [tex]h = h_b -h_a[/tex]
[tex]PE_B = m_b g(h_b - h_a)[/tex]
[tex]KE _B[/tex] is the kinetic energy at B which is 0 (at the top )
Where is [tex]E_F[/tex] is the workdone against velocity which from the diagram is
[tex]\mu_k m_b g cos 25 *d[/tex]
So
[tex]\frac{1}{2} m_b v_2 f^2 = m_b g h_b + \mu_k m_b g cos \25 * d[/tex]
Substituting values
[tex]\frac{1}{2} * 20 * 2.310^2 = 20 * 9.8 * d sin(25) + 0.5* 20 * 9.8 * cos 25 * d[/tex]
So
[tex]d = 0.313 \ m[/tex]
