A 0.5-kg block slides along a horizontal frictionless surface at 2 m/s. It is brought to rest by compressing a very long spring of spring constant 800 N/m. The maximum spring compression is
Answers
Explanation:
As the track AB is frictionless, the block moves this distance without loss in its initial KE=
2
1
mv
2
=
2
1
×0.5×3
2
=2.25J. In the path BD as friction is present, so work done against friction
=μ
k
mgs=0.2×0.5×10×2.14=2.14J
So, at D the KE of the block is=2.25-2.14=0.11J
Now, if the spring is compressed by x
0.11=
2
1
×k×x
2
+μ
k
mgx
i.e., 0.11=
2
1
×2×x
2
+0.2×0.5×10x
or x
2
x−0.11=0
which on solving gives positive value of x=0.1 m
After moving the distance x=0.1 m the block comes to rest. Now the compressed spring exerts a force:
F=kx=2×0.1=0.2N
On the block while limiting frictional force between block and track is f
L
=μ
s
mg
=0.22×0.5×10=1.1N.Since,F<f
L
. The block will not move back. So, the total distance moved by the block
=AB+BD+0.1
=2+2.14+0.1
=4.24m
Answer:
As the track AB is frictionless, the block moves this distance without loss in its initial KE=
2
1
mv
2
=
2
1
×0.5×3
2
=2.25J. In the path BD as friction is present, so work done against friction
=μ
k
mgs=0.2×0.5×10×2.14=2.14J
So, at D the KE of the block is=2.25-2.14=0.11J
Now, if the spring is compressed by x
0.11=
2
1
×k×x
2
+μ
k
mgx
i.e., 0.11=
2
1
×2×x
2
+0.2×0.5×10x
or x
2
x−0.11=0
which on solving gives positive value of x=0.1 m
After moving the distance x=0.1 m the block comes to rest. Now the compressed spring exerts a force:
F=kx=2×0.1=0.2N
On the block while limiting frictional force between block and track is f
L
=μ
s
mg
=0.22×0.5×10=1.1N.Since,F<f
L
. The block will not move back. So, the total distance moved by the block
=AB+BD+0.1
=2+2.14+0.1
=4.24m
hope it help you.