Question

A block m=18.0kg is on a smooth horizontal surface and is connected by a thin cord that passes over a pulley to a second block m=2.0kg which hangs vertically. Find the acceleration of the system.

a=(Fnet)/(ma+mb)=(mbg)/(ma+mb)=[(2)(9.8)]/(18+2) = 0.980 m/s^2

Can someone explain this solution to me?

Thanks :)

Answer 1

I just need someone to explain the solution to me and how they got it

Answer 2

acording to newton's second law of motion the net external force applied on a system is equal to its rate of change of momentum.simplifying this law we get,
net force(F)=mass of the system x acceleration
in this case the net force on the system is the force of gravity=mg
on the small block.
now u have the net force and the total mass of both the blocks is known to u .u can find the acceleration of the two blocks

Answer 3

but why add the masses?

Answer 4

like they did

Answer 5

because the force of gravity is pulling both the blocks as they ar econnected by spring.

Answer 6

and the system here consists of both the blocks

Answer 7

they get F(net) = (Mb)(g)

I'm assuming that F(net) is the unbalanced force??? because all other forces cancel each other out

Answer 8

yes.u r right

Answer 9

did u get the whole solution now???????@Fellowroot

Answer 10

i think so thanks!

Answer 11

Yes, the force of gravity on the 2 kg block has to accelerate
the mass of 2 + 18 = 20 kg of the two blocks.

F=m g = (m + M) a