A 50 kg box rests on a smooth frictionless table. Connected to the box is a light string passing through a pulley at the end of the table. Suspended on rhe other end of the string is a 20 kg box. Find

Question

a) acceleration of the system,
b) the tension in the string
c) the displacement 15 seconds after starting from rest.

Thank you.

Answer 1

F = 20 * 9.8

m = 50 + 20 = 70

a = F/m = 2.8 m/s^2

force on falling 20 kg mass
= 20 (9.8) - T
= m a = 20(2.8)
so T = 140 Newtons

d = (1/2) a t^2
= (1/2) 2.8 (225)
= 315
are you sure you did not mean 1.5 seconds?

Answer 2

Hi Damon. Thank you. It is written on my book, it's really 15 seconds.

Answer 3

big table :)

Answer 4

The table is three football fields long and three football fields high?
I bet it was 0.15 seconds actually :)

Answer 5

Okay then. Maybe it's a typo. I'll ask my teacher about this tom :) uhmm anyway, how can you get the acceleration of the two boxes and tension in the cord if you just have 53 degrees and 30 degrees?

Answer 6

I do not understand this geometry. Is the big one trying to drop down a 53 deg slope and the little one being dragged up a 30 degree slope or what?

Answer 7

Still physics. It has a figure shown here. Can't show u the picture. But its like, theres a box and another box suspended by a cord on either sides of a pulley creating 30 deg and 50 deg.

Answer 8

All I can do is tell you the method.

call M1 the mass of the one that is on string at 50 degrees from vertical (aI assume from vertical
Then gravity force down slope is
M1 g cos 50
T is up slope
net force down is (M1 g cos 50 -T)

same deal for M2
net force down is (M2 g cos 30 -T) 30

difference in those two forces = net force = (M1+M2)a

moreover F = m a for each of the two masses