@Abhisar The diagram shows a 0.74-kg square block at four positions along a track. The track is frictionless between points A and B. Elsewhere, the coefficient of static friction on the block is 0.36. Initially, the block is held in place against a spring by a horizontal external force of magnitude 29 N. Then the external force is removed, and the block is projected forward with speed . The block descends a ramp and has speed at the bottom. At point C, its speed is . The block stops at point D. What is the initial compression of the spring?
Answers What is the initial compression of the spring? 7.3 cm 9.2 cm 2.0 cm 3.7 cm 5.5 cm
V1=1.2m/s V2=1.9m/s V3=1.4m/s
@Abhisar Can u help
The velocities are given and the only thing to find out is compression in the spring. right ?
Yes
What is the (kinetic) energy of the block when it is projected ?
29 n
No! That is the force which compresses the spring.
10.44 I did .36*29
No! I am asking kinetic energy when the block is projected. Don't consider that track with friction now.
I did mass times velocity squared times .5 and got .5328. Not sure how else to solve
Ok. That's correct. Let's use symbols for mass and velocity. Once we get the final result, we shall plug in the given numerical values.
Let the mass of the block be m. Then kinetic energy of the when it is released is \(mv_1 ^2 / 2\)
Where does the block gain this energy from ?
The spring compressed
Yes
What is the energy of spring compression ?
The .5328
Is there any formula to find spring energy ?
.5*k*s^2
Correct! But s in the diagram represents the length of the track with friction. Let's say x is the initial compression. Then spring energy is .5*k*x^2. What is relationship between the restoring force and the spring constant ?
They r equal. What is the value of k
Restoring force, F = -kx So, spring energy = \(0.5\times |F| \times x\) Right ?
Right is f the .5328
F is 29 newton. Now plug in the values and find x.
So x 14.5 bc .5*29
14.5 isn't an answer though
x should be 0.0367 meter
Thanks
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