Question

A furniture removalist applies a 886.1 N force vertically upward to lift a 43 kg box. What is the resultant NET force acting on the box?
(If you could tell me the working out, that would be helpful!)

Answer 1

Answer:

Explanation:

=> It is given that,

Force applied on box, F = 130 N

Friction force between the box and the floor, Ff = 30 N

Time period, t = 5 sec

(a) Net force:

Fnet = F - Ff

= 130 - 30

= 100 N

(b) ) Acceleration acting upon the box, f:

if its starts from rest and attains certain KE after being pushed 25m then,

Initial velocity, u = 0

Displacement, s = 25 m

thus, displacement, s = ut + 1/2 ft²

∴ 25 = 0* 5 + 1/2 * f * (5)²

25 = 0 + 1/2 * f * 25

f = 2 ms⁻²

(c)  Mass of the box, M :

Mass = Force / acceleration

=130/2

= 65 kg

(d) Kinetic energy and final velocity :

=> final velocity, v:

v² = u² + 2fs

v² = (0)² + 2 * 2 * 25

v² = 100

v = 10 ms⁻¹

=> kinetic energy, KE:

KE = 1/2 mv²

     = 1/2 * 65 * (10)²

     = 1/2 * 65 * 100

     = 3250 J

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Answer 2

Answer:

Resultant Net force acting on box is 464.27N

Explanation:

Given that,

Force applied by furniture removalist=886.1N

It acts in vertically upward direction to lift box

Also given,

Mass of box=43kg

Therefore weight of box acting downward is found as follows

$W=mg =43 \times 9.81 = 421.83N$

A force of 886.1N acts in upward direction and weight 421.83N acts in downward direction.Therefore net force is

$F _{net} = 886.1N - 421.83N \\ = 464.27N$

Therefore,Resultant Net force acting on box is 464.27N

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