Does anyone know how to do this problem about force???

Question

A 3.9 kg bucket of water is raised from a well by a rope. If the upward acceleration of the bucket is 2.8 m/s2, find the force exerted by the rope on the bucket.

Answer 1

The net force acting on the bucket by Newton's Second law =
3.9X2.8 = 10.92 N upwards. Forces acting on the bucket are 3.9x9.8 = 38.22 N downwards by the earth and F, the force acted upon by the rope on the bucket which is upwards. So we have

F - 38.22 = 10.92 or F = 38.22+10.92 = 49.14 N is the final answer.

Answer 2

F=39.2/8=109/2

Answer 3

Force
=mass*acceleration
=3.9kg*2.8m/sec^2
=10.92 Newtons

Answer 4

The equation will be like this:

T-mg=ma
where,
T=tension in rope; g=acc. due to gravity; m=mass; a==acc. upwards.
substitute the values of m and a in this equation to get T which is the net force exerted by bucket on rope

Answer 5

Sum Fx = 0 and Sum Fy = 0 Sum Fx = F1,x + F2,x + F3,x = 0 + forty 4 cos 60 + F3,x = 22 + F3,x = 0. So, F3,x = -22 N. Sum Fy = F1,y + F2,y + F3,y = 33 + forty 4 sin 60 + F3,y = 33 + forty 4 sin 60 + F3,y = 0 So, F3,y = -seventy one N. F3 = sqrt [ F3,x ^2 + F3,y ^2 ] = sqrt [ (-22)^2 + (-seventy one)^2 ] = seventy 4 N. process F3 = one hundred eighty + tan^(-one million) (seventy one/22) = one hundred eighty + seventy 3 = 253 deg.

Answer 6

F=M*A. In this case you have A1=g, A2=2.8 m/s^2 and A=their sum. I'm sure you can do that math.

Answer 7

nioton=F=ma m=weight and u know a
F=39.2/8=109/2

Answer 8

i think you take 3.9 and mutiply it by 2.8 (that should give you 10.92) then multiply it by 2(then you will get 21.84).