Question

A 5.0 kg bucket of water is raised from a well by a rope. If the upward acceleration of the bucket is 3.0 m/s^2, find the force exerted by the rope on the bucket..

Answer 1

newtons second law tells you

sum of forces = ma

the forces acting on the bucket are the force of the rope up and gravity down, these combine to create an upward accel of 3 m/s/s

so we have:

sum of forces = F-mg = ma

F-5g=5x3m/s/s

F=5x9.8+5x3 = 49+15=64N

2) F=ma

a=change in speed/time = (5m/s-20m/s)/4s
=-3.75 m/s/s

F=-3.75 m/s/s x 2000kg = -7500N (the minus sign shows it is opposing the motion)

vf^2=v0^2+2ad where vf=final speed (5m/s), vo is initial speed (20 m/s), a is accel (-3.75 m/s/s) and d is the distance to be found:

d=(vf^2-v0^2)/(2a)
d=(5^2-20^2)/(-2x3.75) = 50 m

Answer 2

we know that m = 5 kg
a = 3.0 m/s^2
g = 9.8 (since not written, lets assume it's an international standard)

T - mg = ma

T = mg + ma

T = (5 . 9.8) + (5 . 3 )

T = 49 + 15

T = 64 N

Hope this helps