1) We know that the work done by the spring on the block is given by:
W = (1/2) * k * x1^2
Let's plug in the given values:
W = (1/2) * 4545 * (0.515)^2
W = 605.39 J
So 605.39 J of work is done by the spring as it accelerates the block.
2) We can use the work-energy principle, which states:
W = (1/2) * m * v^2
We have W = 605.39 J and m = 14 kg, so we can solve for v:
605.39 = (1/2) * 14 * v^2
v^2 = 605.39 * 2 / 14
v^2 = 86.482
v = sqrt(86.482)
v = 9.299 m/s
The speed of the block right after it leaves the spring is 9.299 m/s.
3) The work done by friction as the block crosses the rough spot can be calculated using the formula:
W_fric = -μk * m * g * d
We have μk = 0.4, m = 14 kg, g = 9.81 m/s^2, and d = 2.8 m:
W_fric = -0.4 * 14 * 9.81 * 2.8
W_fric = -153.808 J
So friction does -153.808 J of work on the block as it crosses the rough spot.
4) To find the speed of the block after it passes the rough spot, we can use the work-energy principle again:
W_total = (1/2) * m * (v_final^2 - v_initial^2)
We know that W_total = 605.39 J - 153.808 J = 451.582 J, and v_initial = 9.299 m/s:
451.582 = (1/2) * 14 * (v_final^2 - 9.299^2)
v_final^2 = 451.582 * 2 / 14 + 9.299^2
v_final^2 = 64.512 + 86.482
v_final^2 = 150.994
v_final = sqrt(150.994)
v_final = 12.287 m/s
The speed of the block after it passes the rough spot is 12.287 m/s.
5) To find how far into the rough path the block slides before coming to rest, we can use conservation of energy. We can write:
(1/2) * k * x2^2 = Π * m * g * d
Plugging in the values:
(1/2) * 4545 * (0.151)^2 = 3.14 * 14 * 9.81 * d
45.76 = 431.572 * d
d = 45.76 / 431.572
d ≈ 0.106 m
The block slides 0.106 m into the rough patch before coming to rest.
6) We'll find the distance the spring needs to be compressed so that the block will just barely make it past the rough patch when released. For this, we can equate the work done by the spring to the work done against friction:
(1/2) * k * x^2 = μk * m * g * (d+2.8)
Plugging in the values:
(1/2) * 4545 * x^2 = 0.4 * 14 * 9.81 * (2.8+d)
2272.5x^2 = 43.542 * (2.8+d)
x^2 = 43.542 / 2272.5 * (2.8+d)
x^2 = 0.01916 * (2.8+d)
Now we need to solve for d:
0.01916 * (2.8 + d) = (0.515)^2
d ≈ 0
For this particular problem, the block will barely make it past the rough patch when the spring is compressed the entire 0.515 m.
A block with mass m = 14 kg rests on a frictionless table and is accelerated by a spring with spring constant k = 4545 N/m after being compressed a distance x1 = 0.515 m from the spring’s unstretched length. The floor is frictionless except for a rough patch a distance d = 2.8 m long. For this rough path, the coefficient of friction is ìk = 0.4.
How much work is done by the spring as it accelerates the block?
What is the speed of the block right after it leaves the spring?
How much work is done by friction as the block crosses the rough spot?
What is the speed of the block after it passes the rough spot?
Instead, the spring is only compressed a distance x2 = 0.151 m before being released.
How far into the rough path does the block slide before coming to rest?
What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released?
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