Part A. A 66.0-kg box is pulled by a rope making an angle of = 21.5° with the ho
ID: 1332074 • Letter: P
Question
Part A.
A 66.0-kg box is pulled by a rope making an angle of = 21.5° with the horizontal. The tension in the rope is
25.5 N.(Assume the surface the box slides across is frictionless.)
If the box starts from rest, determine the time required to move the box a horizontal distance of 2.35 m.
Part B.
The figure shows two blocks connected by a lightweight string across a frictionless pulley. The hanging block has mass m1 = 5.75 kg and the second block has mass m2 = 3.80 kg.The ramp is frictionless and inclined at an angle of = 43.5° from the horizontal. When released from rest, block 1 will descend and block 2 will move up the ramp. (Assume the ramp is frictionless.)
Determine the magnitude of each block's acceleration and the tension in the string.
a=
T=
Explanation / Answer
Tx = T*cos21.5
ax = Tx/m = (25.5*cos21.5)/66 = 0.36 m/s^2
along horizantal
x = vox*t + 0.5*ax*t^2
2.35 = 0 + 0.5*0.36*t^2
t = 3.61 s <<<--answer
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part(B)
for m1
Fg = m1*g (down ward)
tension = T in upwards
m1 is moving down wards net force is down wards
-m1*g + T = -m1*a
m1*g - T = m1*a ....(1)
for m2
component of weight down the incline = m*g*sintheta
Fnet = T - m2*g*sintheta
from newtons 2nd law net force = m2*a
T - m2*g*sintheta = m2*a ........(2)
1 + 2
m1*g - m2*g*sintheta = a*(m1+m2)
a = (m1*g - m2*g*sintheta)/(m1+m2)
a = ((5.75*9.8)-(3.8*9.8*sin43.5))/(5.75+3.8)
a = 3.22 m/s^2
T = m1*g - m1*a
T = 5.75*(9.8-3.22)
T = 37.8 N <<<----------------answer