Answer:
a) 12.9 m/s
b) 11.3 m/s
c) 9.3 m
Explanation:
Given that
Speed o the block, v(o) = 7 m/s
Length of the section, l = 12 m
Coefficient of kinetic friction, μ = 0.7
Heights h1 and h2 = 6 m and 2 m
v² = v(o)² + 2gh
v² = 7² + 2 * 9.8 * 6
v² = 49 + 117.6
v² = 166.6
v = √166.6
v = 12.91 m/s
v² = v(o)² + 2g(h1 - h2)
v² = 7² + 2 * 9.8 * 4
v² = 49 + 78.4
v² = 127.4
v = √127.4
v = 11.29 m/s
k = 1/2mv²
k = 1/2 * m * 11.29²
k = 63.73m
k = (μmg)d
63.73m = 0.7 * m * 9.8 * d
d = 63.73 / 6.86
d = 9.29 m
The block doesn't reach D. A further explanation is below.
(A)
By using the law of conservation of energy, we get
→ [tex]Energy \ at \ A = Energy \ at \ B[/tex]
→ [tex](0.5\times m\times u^2)+(m\times g\times h_1) = 0.5\times m\times v^2[/tex]
[tex](0.5\times 7^2)+(9.8\times 6) = 0.5\times v^2[/tex]
[tex]v = 12.9 \ m/sec[/tex] (Speed at B)
(B)
→ [tex]Energy \ at \ B = Energy \ of \ C[/tex]
→ [tex](0.5\times m\times v^2) = (m\times g\times h_2)+(0.5\times m\times v_2^2)[/tex]
[tex]0.5\times 12.9^2 = (9.8\times 2)+(0.5\times v_2^2)[/tex]
[tex]v_2 = 11.3 \ m/sec[/tex] (Speed at C)
(C)
Work done by the frictional force = Change in K.E
→ [tex]-fk\times L = -0.5\times m\times v_2^2[/tex]
[tex]mu_k\times g\times m\times L = 0.5\times m\times v_2^2[/tex]
[tex]0.7\times 9.8\times L=0.5\times 11.3^2[/tex]
[tex]L = 9.3 \ m[/tex]
∴ 9.3 m < 12 m
Thus the above answer is correct.
Learn more:
https://brainly.com/question/13199734
