A few days ago
Please tell me my error in this intro physics prob?
A relaxed spring with spring constant k = 70 N/m is stretched a distance di = 66 cm and held there. A block of mass M = 10 kg is attached to the spring. The spring is then released from rest and contracts, dragging the block across a rough horizontal floor until it stops without passing through the relaxed position, at which point the spring is stretched by an amount df = di/7
7What is the coefficient of kinetic friction µk between the block and the floor?
heres what i did, with the idea mu-max is when opposing forces are equal;
Fsp = -K delta x
Fk = mu N
so i equated equations
-K delta x = mu N
mu = -K delta s / N
mu = -70(.094) / (10*9.8)
mu = .067
=wrong????
Top 3 Answers
A few days ago
Favorite Answer
The value seems right. Why do you feel it is wrong?
The residual force on the spring=(70/9.8)*(66/7)/100. Since the mass has just stopped moving this must be=mu*10.
Equating the two and solving for mu, one finds that mu=0.067
0
4 years ago
Assuming the slope ? is 30° to the horizontal, The tension performing parallel to the slope Fs = m*g*sin?. it is the only tension inflicting acceleration. considering the fact that a = F/m, a = mgsin?/m = g*sin? = 9.8*.5 = 4.9 m/s². observe that the mass isn’t mandatory to compute this acceleration. a similar freebody diagram will teach that the conventional tension Fn = mg*cos? = ninety*9.8*cos30° = 763.eighty 3 N
0
A few days ago
Just move the block yourself
0
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