1. In a control system, an accelerometer consists of a 4.87-g object sliding on
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Question
1. In a control system, an accelerometer consists of a 4.87-g object sliding on a calibrated horizontal rail. A low-mass spring attaches the object to a flange at one end of the rail. Grease on the rail makes static friction negligible, but rapidly damps out vibrations of the sliding object. When subject to a steady acceleration of 0.784g, the object should be at a location 0.460 cm away from its equilibrium position. Find the force constant of the spring required for the calibration to be correct.
N/M ????
2. You can think of the work?kinetic energy theorem as a second theory of motion, parallel to Newton's laws in describing how outside influences affect the motion of an object. In this problem, solve parts (a), (b), and (c) separately from parts (d) and (e) so you can compare the predictions of the two theories. A 16.0-g bullet is accelerated from rest to a speed of 730 m/s in a rifle barrel of length 71.8 cm.
(a) Find the kinetic energy of the bullet as it bullet as it leaves the barrel.
(b) Use the work-kinetic energy theorem to find the net work that is done on the bullet.
(c) Use the result to part (b) to find the magnitude of the average net force that acted on the bullet while it was in the barrel.
(d) Now model the bullet as a particle under constant acceleration. Find the constant acceleration of the bullet that starts from rest and gains a speed of 730 m/s over a distance of 71.8 cm.
(e) Modeling the bullet as a particle under a net force, find the net force that acted on it during its acceleration.
(f) What conclusion can you draw from comparing your results of parts (c) and (e)
3.
A block of mass m = 6.40 kg is released from rest from point A and slides on the frictionless track shown in the figure below. (Let ha = 6.70 m.)
Determine the block's speed at points B and C
b) Determine the net work done by the gravitational force on the block as it moves from point B to point C.
Explanation / Answer
1) F = k*x
m*a = k*x
==> k = m*a/x = 4.87*10^-3*0.784*9.8/(0.46*10^-2) = 8.13 N/m
2) m = 16*10^-3 kg,
v1 = 0, v2 = 730 m/s, d = 0.718 m
a) KE2 = 0.5*m*v2^2 = 4263.2 J
b) workdone, W = chnage in kinetic energy = KE2 - KE1 = 4263.2 - 0 = 4263.2 J
c) W = Fav*d
==> Fav = w/d = 4263.2/0.718 = 5937.6 N
d)
v2^-2-v1^2 = 2*a*d
==> a = (v2^2-v1^2)/(2*d) = 371100 m/s^2
e) Fav = m*a = 5937.6 N
f) in both cases we get the same result.
3)
h = 6.7 m
let v is the velocity of ball at bottom point
0.5*m*v^2 = m*g*h
==> v = sqrt(2*g*h) = 11.46 m/s
b) workdone = m*g(hb-hc)
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