other previous submissions, also incorrect: c) 6.85, 2.88, 6.947, 4.78 d) 0.698,
ID: 1597314 • Letter: O
Question
other previous submissions, also incorrect:
c) 6.85, 2.88, 6.947, 4.78
d) 0.698, 2.88, 6.947, 0.487
We show below a free-body diagram of a 1.00-kg block pulled across a surface by a string. The string's tension is T = 6 N. The string makes an angle of 37 degrees with the horizontal. The coefficient of kinetic friction is u_k = 0.30. Determine the y-component of the normal force. Determine the x-component of the force of friction. ? Find the magnitude of the total force on the block. Finally, find the magnitude of the acceleration of the block.Explanation / Answer
T = 6 N , uk = 0.3 , m = 1kg
(a) Net force along vertical direction
FN + Tsin(theta) = mg
FN +6*sin(37)= 1*9.8
FN = 6.19 N
(b) FK = uk*FN
FK = 0.3*6.19
FK = 1.86 N
(c) from Newton second law
F = -Fk +Tcos(theta)
F = -1.86 + 6*cos(37)
F = 2.93 N
(d) Net force along horizotnal direction
F = ma
2.93 = 1*a
a = 2.93 m/s^2
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