please explain it in English
Answers
Answer:
We refer to its starting point as A, the point where it first comes into contact with the spring as B, and the point where the spring is compressed x
0
=0.55m, as C, as shown in the figure below. Point C is our reference point for computing gravitational potential energy. Elastic potential energy (of the spring) is zero when the spring is relaxed.
Information given in the second sentence allows us to compute the spring constant. From Hooke's law, we find
k=
x
F
=
0.02m
270N
=1.35×10
4
N/m.
The distance between points A and B is l
0
and we note that the total sliding distance l
0
+x
0
is related to the initial height h
A
of the block (measured relative to C) by
sinθ=
l
0
+x
0
h
A
where the incline angle θ is 30
∘
.
(a) Mechanical energy conservation leads to
K
A
+U
A
=K
C
+U
C
⇒0+mgh
A
=
2
1
kx
0
2
which yields
h
A
=
2mg
kx
0
2
=
2(12kg)(9.9m/s
2
)
(1.35×10
4
N/m)(0.055m)
2
=0.174m.
Therefore, the total distance traveled by the block before coming to a stop is
l
0
+x
0
=
sin30
∘
h
=
sin30
∘
0.174m
=0.347m≈0.35m.
(b) From this result, we find l
0
=x
0
=0.347m−0.055m=0.292m, which means that the block has descended a vertical distance
∣Δy∣=h
A
−h
B
=l
0
sinθ=(0.292m)sin30
∘
=0.146m
in sliding from point A to point B. Thus, using Eq. 8−18, we have
0+mgh
A
=
2
1
mv
B
2
+mgh
B
⇒
2
1
mv
B
2
=mg∣Δy∣
which yields v
B
=
2g∣Δy∣
=
2(9.8m/s
2
)(0.146m)
=1.69m/s≈1.7m/s.
Note: Energy is conserved in the process. The total energy of the block at position B is
E
B
=
2
1
mv
B
2
+mgh
B
=
2
1
(12kg)(1.69m/s)
2
+(12kg)(9.8m/s
2
)(0.028m)=20.4J,
which is equal to the elastic potential energy in the spring:
2
1
kx
0
2
=
2
1
(1.35×10
4
N/m)(0.55m)
2
=20.4J.