A 0.50 kg air-track glider has an initial speed of 0.25 m/s as it passes through
ID: 1477383 • Letter: A
Question
A 0.50 kg air-track glider has an initial speed of 0.25 m/s as it passes through a photoelectric gate that starts a timer. As it passes through a constant force of 0.40 N is applied in the direction of motion. (a) What is the acceleration of the glider? (b) The glider then passes through a second gate that stops the timer at 1.3 s. What is the distance between the two gates? (c) The 0.40 N force is applied by means of a string attached to the glider. The other end of the string passes over a frictionless pulley and is attached to a hanging mass m. How big is the mass m? (d) Derive an expression for the tension T in the string as a function of the mass M of the glider, the mass m of the hanging mass, and the acceleration of gravity g.
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Explanation / Answer
Force=0,40 N
We know that force=ma
where, m=mass
a=acceleration
S0,0.40=0.50*a
a=0.8 m/s2
b) time=1.3 s
velocity=0.25 m/s
S0, we know that velocity=distance/time
distance=velocity*time=0.25 * 1.3=0.325 m
c. The 0.40N force must be constant throughout the string. Therefore, the NET acceleration of the mass (= accel of glider necessarily) must produce a force in the mass of 0.40N
If a is the acceleration you calculated and g is gravity, then
m = F / (g - a) = 0.40N / (9.8m/s2 - a)
m=0.40/(9.8-0.8)=0.044 kg
4. Tension in the string = M * a
(that is, the mass of the glider times its acceleration). Therefore
a = T / M
Tension in the string is also = m(g - a)
(that is, the hanging mass * the net acceleration). Subbing for a:
T = m(g - T/M) = mg - (m/M)T
T + (m/M)T = mg
T(1 + m/M) = mg
T = mg / (1 + m/M)
which can be restated a number of ways, including
T = mMg / (M + m)
T = g / ( 1/m + 1/M)
T = Mg / ( M/m + 1)
As M gets very large relative to m, T ---------> mg
As M gets very small relative to m, T --------> 0
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