Can you please help me understand question 1 & 2 step-by-step 1) A mass is attac
ID: 1564011 • Letter: C
Question
Can you please help me understand question 1 & 2 step-by-step
1) A mass is attached to the end of a string and spun in a vertical circle of radius r. The rope breaks when the mass is traveling directly upwards. The mass rises to a height h above the point it was when the string broke. (At point shown in diagram below.) Let: mass m = 175 grams, radius r = 170 cm, and height h = 10.0 meters What was the tension in the string when it broke?
2) A 3.622 kg box is dragged over a rough horizontal surface by a constant force of 16 N acting at an angle of 37° above the horizontal as shown above. The speed of the box increases from 5.0 m/s to 6.0 m/s in a displacement of 775 centimeters. The coefficient of kinetic friction is about
Explanation / Answer
Given
1)
mass m = 175 g = 0.175 kg, r = 170 cm = 1.7 m, h = 10 m
the speed of the mass when the sting broke is v = sqrt(2gh)
v = sqrt(29.8*10) m/s = 17.26 m/s
the tension in the string is centripetal force F = mv^2/R N = T
T = mv^2/R = 0.175*17.26^2/1.7 N
T = 30.67 N
2) mass of box m = 3.622 kg, constant force applied F = 16 N
then angle is theta = 37 degrees above the horizontal
change in speed is dV = 1 m/s
displacement s = 775 m = 7.75 m
from work energy theoerem total work done = change in kinetic energy
work done = (F cos theta - mue_k*mg cos theta)(s)
= (16cos37 - mue_k*3.622*9.8 cos 37)
change in k.e = 0.5(3.662)(1)^2 J = 1.831 J
equating these two equations
(16cos37 - mue_k*3.622*9.8 cos 37) = 1.831
mue_k = 0.3861
the coefficient of kinetic frictiono mue_k= 0.3861