1. The figure below shows a block of mass 2.5 kg being pushed up an ideal inclin
ID: 1282654 • Letter: 1
Question
1. The figure below shows a block of mass 2.5 kg being pushed up an ideal incline. The block moves up the incline with an acceleration a = 0.75 [m/s/s]. Calculate the pushing force that was applied to the block.
2. The figure below shows a heavy block of mass 15 kg being pulled up a loading ramp. The ramp is sloped at an angle of 20 degrees with the horizontal and a pulling force of magnitude 300 N is applied in a direction which makes an angle of 30 degrees with the ramp. Calculate the acceleration of the block
3. The figure below shows a block of mass 1.5 kg sliding down an ideal incline. The incline is followed by an ideal horizontal floor 20 cm in length, which is in turn followed by a second ideal incline. The block is released from rest. By dividing the motion of the block into three segments, calculate the time the block spends in each segment
(a) time spent in segment I
(b) time spent in segment II
(c) time spent in segment III
1. The figure below shows a block of mass 2.5 kg being pushed up an ideal incline. The block moves up the incline with an acceleration a = 0.75 [m/s/s]. Calculate the pushing force that was applied to the block. 3. The figure below shows a block of mass 1.5 kg sliding down an ideal incline. The incline is followed by an ideal horizontal floor 20 cm in length, which is in turn followed by a second ideal incline. The block is released from rest. By dividing the motion of the block into three segments, calculate the time the block spends in each segment (a) time spent in segment I (b) time spent in segment II (c) time spent in segment III 2. The figure below shows a heavy block of mass 15 kg being pulled up a loading ramp. The ramp is sloped at an angle of 20 degrees with the horizontal and a pulling force of magnitude 300 N is applied in a direction which makes an angle of 30 degrees with the ramp. Calculate the acceleration of the blockExplanation / Answer
1)
F_push = m*g*sin(35) + m*a
= 2.5*9.8*sin(35) + 2.5*0.75
= 15.93 N
2)
a)
a) d = h/sin(50) = 30/sin(50) = 39.16 cm = 0.3916 m
a = g*sin(50) = 9.8*sin(50) = 7.5 m/s^2
d = u*t + 0.5*a*t^2
d = 0 + 0.5*a*t^2
t = sqrt(2*d/a)
= sqrt(2*0.3916/7.5)
= 0.323 s
b) v = sqrt(2*9.8*0.3) = 2.42 m/s
t = d/v = 0.2/2.42 = 0.082 s
c)
c) a = -g*sin(35)
= -9.8*sin(35)
= -5.62 m/s^2
v = u + a*t
t = (v-u)/a
= (0-2.42)/(-5.62)
= 0.43 s
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