I have solved almost of this problem except the last 2 parts. 1) How much work i
ID: 1439963 • Letter: I
Question
I have solved almost of this problem except the last 2 parts.
1)
How much work is done by the spring as it accelerates the block?
534.49m/s
2- What is the speed of the block right after it leaves the spring?
8.73m/s
3-
3)
How much work is done by friction as the block crosses the rough spot?
-140.63 J
4-What is the speed of the block after it passes the rough spot?
7.49 m/s
5-Instead, the spring is only compressed a distance x2 = 0.127 m before being released.
How far into the rough path does the block slide before coming to rest?
( I need help on Part 5) .. The answer should be on (m)
6-What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released?
(I need help on part 6) .. The aswer should be on (m)
Explanation / Answer
1) Work donr by spring = 0.5kx^2 = 0.5 * 5279 * 0.45^2 = 534.49 J
This energy is transfered to the block as the spring expands.
2) 0.5 * 14 * v^2 = 0.5 * 5279 * 0.45^2
v = (5279 * 0.45^2 ) / 14)^0.5 = 8.73 m/s
3) Friction work = u * m * g * d = (0.41 * 14 * 9.8 * 2.5) = 140.63 J
4) The work done by the the friction force reduces the kinetic energy of the block.
Final KE = Initial KE - Work done by friction
0.5 * 14 * v^2 = (0.5 * 5279 * 0.45^2 ) - (0.41 * 14 * 9.8 * 2.5)
7*v^2 = 534.49 - 140.63
v = 7.5 m/s
5) If the spring is compressed 0.127 m, the work done by the spring is 0.5 * 5279 * 0.45^2 . So the initial KF of the block = 0.5 * 5279 * 0.127^2. When the block comes to rest, the KE = 0.
The work done by the friction force must decrease the kinetic energy of the block from (0.5 * 5279 * 0.127^2) to 0.
0.41* 14 * 9.8 * ( length of rough path) = (0.5 * 5279 * 0.127^2)
( length of rough path) = (0.5 *5279 * 0.127^2) / (0.41* 14 * 9.8) = 0.757 m
6) The work done by the spring must be "barely" greater than than the worjk done by the friction!
0.5 * k * x^2 = u * m* g * d
x^2 = (u * m* g * d) / (0.5 * k) = (0.41 * 14 * 9.8 * 0.757) / (0.5 * 5279) = 0.01613
x = 0.127 m
If the the spring is compressed barely more than 0.127 meter, the block will come to rest at a point barely past the 0.757 m long rough patch!
If the rough patch is the original 2.5 m long:
x^2 = (u * m* g * d) / (0.5 * k) = (0.41 * 14 * 9.8 * 2.5) / (0.5 * 5279) =
x = 0.231 m