![hiiamsuperverylong](/avatars/46243.jpg)
hiiamsuperverylong
08.02.2021 •
Physics
A light spring of force constant 3.85 N/m is compressed by 8.00 cm and held between a 0.250 kg block on the left and a 0.500 kg block on the right. Both blocks are at rest on a horizontal surface. The blocks are released simultaneously so that the spring tends to push them apart. What is the maximum velocity each block attains if the coefficient of kinetic friction between each block and the surface is:.
A) 0
B) 0.100
C) 0.463
Assume that the coefficient of static friction is larger than that for kinetic friction.
Solved
Show answers
More tips
- H Health and Medicine Why Wearing a Back Brace Can Be Beneficial During Back Strain?...
- S Sport When and Where Will the 2014 World Cup be Held?...
- C Computers and Internet How to Choose a Monitor?...
- S Style and Beauty How to Get Rid of Peeling Nails: Natural Remedies...
- S Science and Technology Understanding Magnetic Storms: A Guide to This Natural Phenomenon...
- F Family and Home What is Most Important in Men s Lives?...
- G Goods and services Which TV is better - LCD or Plasma?...
- C Computers and Internet Are there special gaming mice?...
- G Goods and services LED-подсветка в LCD-телевизорах: 5 причин, почему она лучше других технологий...
- C Computers and Internet Keep Your Mouse Pad Clean: The Right Way to Clean It?...
Answers on questions: Physics
- M Mathematics 105.25 x 70.75 NO LINKS...
- A Arts Which is an example of a found instrument a)steel drum b) old violin c) flute d) tympani...
- M Mathematics To take advantage of long term gains in the stock market, the best strategy is usually to... avoid the stock market and save money instead. buy and sell stocks frequently....
Ответ:
A) v1 = -0.256 m/s
v2 = 0.128 m/s
B) v1 = -0.0642 m/s
v2 = 0 m/s
C) v1 = v2 = 0 m/s
Explanation:
We are given;
Spring constant; k = 3.85 N/m
Distance compressed; x = 8 cm = 0.08 m
Mass of left block; m1 = 0.25 kg
Mass of right block; m2 = 0.5 kg
A) From conservation of energy;
½kx² = ½m1•v1² + ½m2•(v2)²
Also from conservation of linear momentum, we know that;
m1v1 = m2v2
Now, v2 = m1•v1/m2
Plugging this for v2 in the first equation gives;
½kx² = ½m1•v1² + ½m2•(m1•v1/m2)²
Making v1 the subject, we have;
v1 = √(kx²(m2))/(m1•m2 + (m1)²))
v1 = √[(3.85 × 0.08² × 0.5)/((0.25 × 0.5) + 0.5²))
v1 = 0.256 m/s
This is the left block which is in the negative x direction and so v1 = -0.256 m/s
We saw that v2 = m1•v1/m2
Thus; v2 = (0.25 × 0.256)/0.5
v2 = 0.128 m/s
B) Force exerted by spring is;
F_s = kx = 3.85 × 0.08
F_s = 0.308 N
Normal forces will be calculated for both blocks as;
N1 = m1•g = 0.25 × 9.8 = 2.45 N
N2 = m2•g = 0.5 × 9.8 = 4.9 N
Let's calculate force of static friction for both blocks;
F_s1 = μN1 and F_s2 = μN2
We are given coefficient of friction as
μ = 0.1.
Thus;
F_s1 = 0.1 × 2.45 = 0.245 N
F_s2 = 0.1 × 4.9 = 0.49 N
F_s1 is lesser than F_s. Thus let's calculate the new compression;
x_1 = 0.245/3.85
x_1 = 0.06364 m
Thus, change in compression is;
Δx = x - x_1
Δx = 0.08 - 0.06364
Δx = 0.01636 m
From conservation of energy, since our coefficient of friction is not zero and we have frictional force, then we use the equation;
½k(x)² - F_s1•Δx = ½m1•v1² + ½k(x_1)²
Making v1 the subject, we have;
v_1 = √(k(x² - x_1²) - 2F_s1•Δx)/m1
v_1 = √(3.85(0.08² - 0.06364²) - 2(0.245 × 0.01636)/0.25
v1 = 0.0642 m/s
Since in negative x direction, then
v1 = -0.0642 m/s
F_s2 is greater than F_s. Thus, it means the right side block will not move and velocity is zero. v2 = 0 m/s
C) Coefficient of friction is now 0.463.
Thus;
F_s1 = 0.463 × 2.45 = 1.13435 N
F_s2 = 0.463 × 4.9 = 2.2687 N
They are both greater than F_s and thus no motion in both cases.
So v1 = v2 = 0 m/s
Ответ: