A 6.0kg box is on a frictionless 37? slope and is connected via a massless strin
ID: 1290997 • Letter: A
Question
A 6.0kg box is on a frictionless 37? slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2.1kg weight.
PART A)
What is the tension in the string if the 6.0kg box is held in place, so that it cannot move?
Express your answer using two significant figures.
PART B) If the box is then released, which way will it move on the slope? UPWARD OR DOWNWARD?
PART C)
What is the tension in the string once the box begins to move?
Express your answer using two significant figures.
Explanation / Answer
F = m a
a = 0 m/s2 [The system is not moving]
F = 0
T - m g = 0
T = m g = 2.1 kg * 9.81 m/s2
T = 20.60 N
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upwards
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The system is now moving. The two blocks will have same acceleration and same tension
m1 = 6 kg , m2 = 2.1 kg
m1 g sin (37) - T = m1 a --------------(1)
T- m2 g = m2 a ---------------(2)
Solve for T in eqn (2)
T= m2 g + m2 a
m1 g sin (37) - (m2 g + m2 a) = m1 a
m1 g sin (37) - m2 g - m2 a = m1 a
Solve for a
m1 g sin (37) - m2 g = a * (m1 + m2)
a = [m1 g sin(37) - m2 g] / [m1 + m2]
= [6 * 9.8 sin(37) - 2.1 * 9.8] / [6 + 2.1]
a = 1.83 m/s2
sub use a in equation(2)
T = m2 g + m2 a
T = (2.1 * 9.81 ) + (2.1 * 1.83 )
T = 24.42 N