It is found that a force of 18-N is required to stretch a spring by 0.300 meters
ID: 1338773 • Letter: I
Question
It is found that a force of 18-N is required to stretch a spring by 0.300 meters from its natural length. (a) What is the spring constant of the spring? (b) How much energy is stored in the spring when it is stretched by 0.125 meters from its natural length? (c) If the spring is laid horizontally and a 0.250-kg mass is attached to it while initially stretched by 0.125-m, What speed will the mass have after traveling 0.100-m due to the force exerted by the spring? (Assume the mass is in contact with a horizontal, frictionless surface) Answers: (a) 60 N/m (b) 0.469 J (c) 1.90 m/s
A 2.50-kg block is pushed 2.20-m along a horizontal table by a constant force of 16.0-N directed at 25º below the horizontal. If the coefficient of kinetic friction between the block and table is 0.213, what is the work done by the frictional force answer: -14.7J
Explanation / Answer
1)
a) Apply hook's law.
F = k*x
==> k = F/x
= 18/0.3
= 60 N/m
b) Energy stored, u = 0.5*k*x^2
= 0.5*60*0.125^2
= 0.46875 J
c)
0.5*m*v^2 = 0.5*k*(x1^2 - x2^2)
v = sqrt(k*(x1^2 - x2^2)/m)
= sqrt(60*(0.125^2 - 0.025^2)/0.25)
= 1.897 m/s
2) Workdone by friction, W = Friction*d*cos(180)
= N*mue_k*d*(-1)
= (2.5*9.8 + 16*sin(25))*0.213*2.2*(-1)
= -14.65 J
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.