For the following problems, use a glider mass of 0.346 kg. Atwood machine: A gli

Question

For the following problems, use a glider mass of 0.346 kg. Atwood machine: A glider moves frictionlessly along a level air track, tugged by a vertically suspended mass of 0.050 kg that is attached to the glider with a string that is looped over the pulley (refer Figure 5). What is the expected acceleration of the glider? Sloping air track: A glider moves frictionlessly along an air track that has been tilted by 1.5 inches over a span of 1.00 meters. What is the expected acceleration of the glider?

Explanation / Answer

(a)According to system force equation will be

T-m2g = -m2a

where T is tension equal to T=m1a, m1=.346kg and a is acceleration , m2 = .050kg, g= 9.8m/s^2

putting these value in equation and solving

.346*a-.05*9.8=.05*a

a=1.23m/s^2

(b) For slopping air track tilted angle is given by

tilted height/span = 1.5 inches/1m

converting inches to meter 1.5 *2.54/100 m

Force equation is

ma = mgsin (theta)

a= 9.8 *sin(.0381)

a=.37m/s^2

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