You are riding in an elevator on the way to the eighteenth floor of your dormitory. The elevator's upward acceleration is 1.91 m/s2. Beside you is the box containing your new computer; box and contents have a total mass of 29.0 kg. While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. Part A If the coefficient of kinetic friction between the box and elevator floor is 0.35, what magnitude of force must you apply

F = 80.1 N

Explanation:

The force applied on the box must at least be equal to the frictional force of elevator floor on the box. Therefore,

F = μR equation (1)

where,

F = Force applied on box

μ = coefficient of kinetic friction between box and elevator floor = 0.35

R = Normal Reaction = Weight of box and its contents = W

In this case since, the elevator is moving upward with constant acceleration. Therefore, the weight will be given as:

W = m(g + a)

where,

m = mass of box and contents = 29 kg

g = 9.8 m/s²

a = upward acceleration of elevator = 1.91 m/s²

Therefore,

W = R = (29 kg)(9.8 m/s² - 1.91 m/s²)

R = 228.81 N

Therefore, using values in equation (1), we get:

F = (0.35)(288.81 N)

F = 80.1 N

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation