(a)
m u 2 2 = m v 2 2 + μ k m 1 g l \frac{mu^2}{2}=\frac{mv^2}{2}+\mu_km_1gl 2 m u 2 = 2 m v 2 + μ k m 1 g l
v = 2 m ( m u 2 2 − μ k m 1 g l ) = v=\sqrt{\frac{2}{m}(\frac{mu^2}{2}-\mu_km_1gl)}= v = m 2 ( 2 m u 2 − μ k m 1 g l ) =
2 0.05 ( 0.05 ⋅ 85 8 2 2 − 0.4 ⋅ 6 ⋅ 9.8 ⋅ 0.1 ) = 857.94 \sqrt{\frac{2}{0.05}(\frac{0.05\cdot 858^2}{2}-0.4\cdot 6\cdot 9.8\cdot 0.1)}=857.94 0.05 2 ( 2 0.05 ⋅ 85 8 2 − 0.4 ⋅ 6 ⋅ 9.8 ⋅ 0.1 ) = 857.94 m / s m/s m / s
(b)
there is no energy loss for the deformation of the block 1 there is no conversion of mechanical energy into thermal energy
(c)
m v = ( m 2 + m ) V → V = m v m 2 + m = 0.05 ⋅ 857.94 8 + 0.05 = 5.33 mv=(m_2+m)V \to V=\frac{mv}{m_2+m}=\frac{0.05\cdot 857.94}{8+0.05}=5.33 m v = ( m 2 + m ) V → V = m 2 + m m v = 8 + 0.05 0.05 ⋅ 857.94 = 5.33 m / s m/s m / s
(d)
( m 2 + m ) V 2 2 = k d 2 2 + μ k ( m 2 + m ) g d \frac{(m_2+m)V^2}{2}=\frac{kd^2}{2}+\mu_k(m_2+m)gd 2 ( m 2 + m ) V 2 = 2 k d 2 + μ k ( m 2 + m ) g d
d ≈ 0.231 d\approx0.231 d ≈ 0.231 m m m
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