A 15-kg block is on a ramp which is inclined at 20° above the horizontal. It is

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A 15-kg block is on a ramp which is inclined at 20° above the horizontal. It is connected by a string to a 19-kg mass which hangs over the top edge of the ramp. Assuming that frictional forces may be neglected, what is the downward acceleration of the 19-kg block?



Apr 23rd, 2015

Calculate the forces on both blocks. 

The one that hangs over the ramp is pulled straight down by gravity, so F = m g = 19 * 9.81 = 186.39 N. For the one on the ramp, you need to decompose the forces. The force along the ramp is m g sin( 20° ) = 15 * 9.81 * sin( 20° ) = 50.3 N. This force works in the opposite direction of the other force, so the resulting force is 136.1 N. This force works on both blocks, so the acceleration is a = F / m_tot = 136.1 / (19 + 15) =

=The correct answer is 4.0m/s^2

Apr 23rd, 2015

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Apr 23rd, 2015
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Apr 23rd, 2015
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