A van accelerates down a hill (see figure below), going from
rest to 30.0 m/s in 5.30 s. During the
acceleration, a toy (m = 0.210 kg)
hangs by a string from the van's ceiling. The acceleration is such
that the string remains perpendicular to the ceiling.
a) Determine the angle
b) Determine the tension in the string
-
The average velocity Vavg = 30.0m/s / 2 = 15 m/s
so the distance covered was s = Vavg * t = 15.0m/s * 5.30s = 79.5 m
and s = 79.5 m =
v = u +a*t
30 = 0 +a*5.3
a= 5.66 m/s^2
g*sin(theta) = 5.66
sin(theta) = 5.66/9.81
a) theta = 35.24 degree
b)
T = mg*cos(theta)
= 0.21*9.81*cos(35.24) = 1.682 N
The average velocity Vavg = 30.0m/s / 2 = 15 m/s
so the distance covered was s = Vavg * t = 15.0m/s * 5.30s = 79.5 m
and s = 79.5 m =
1)mgsin(theta)=ma
mgcos(theta)=T
a=(30/5.3)
a=5.67m/s^2
sin(theta)=5.67/9.8
theta=35.28
2)T=mgcos(theta)=0.8*9.8*0.21
T=1.68N
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