1.A 4 kg block moves on a horizontal,friction less surface and collides with a s
ID: 1409556 • Letter: 1
Question
1.A 4 kg block moves on a horizontal,friction less surface and collides with a spring of spring constant k that is fixed a wall. when the block momentarily stops, the spring has been compressed by 0.20 m. After rebounding, the block has a speed of 1.00 m/s. Next, the spring is put on an inclined surface with its lower end fixed in place. The same block is now released on the incline at a distance of 5.00 m from the spring's free end When the block momentarily stops, the spring has been compressed by 0.30 m. (a) What is the coefficient of kinetic friction between the block and the incline? (b) How far does the block then move up the incline from the stopping point?
2fric.A 3.5 kg block is accelerated from rest by a compressed spring of spring constant 640 N/m. The block leaves the spring at the spring's relaxed lenght and then travels over a horizontal floor with a coefficient of kinetic friction 0.25. The frictional force stops the block in distance D=7.8 . What are (a)the increase in the thermal energy of the block-floor system, (b)the maximum kinetuc energy of the block, and (c) the original compression distance of the spring?
Explanation / Answer
2) a) The frictional work is the increase in thermal energy W f = Ff * D = k*m*g*d
W f = k*m*g*d
W f = 0.25*3.5*9.8*7.8
W f = 66.89 J
b) The frictional force stopped the block We can compute the kinetic energy that it had at the beginning of the trip.
K.E = W f
K.E = 66.89 J
c) This KE was potential energy stored in the spring. We can now find the compression of the spring.
K.E = 0.5*k*x^2
66.89 = 0.5*640*x^2
x = 0.457 m
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