You are pushing a large crate from the back of a freight elevator to the front as the elevator is moving. Assume that there is friction between the elevator floor and the crate. In Case A, the elevator is accelerating upward and in Case B, it is accelerating downward. Does it require more force for you to move the crate in Case A, Case B, or is the amount of force in each case the same

Case A requires more force than Case B

Explanation:

For this exercise, we propose Newton's second law, we assume that the box is pushed on the x-axis

X axis

    F -fr = m aₓ

    F = fr + m aₓ

Y Axis

     N- W = m

     N = m a_{y} + m g

     N = m (a_{y} + g)

The friction force has as an expression

     fr = μ N

We substitute

    F = μ m (g + a_{y}) + m aₓ

Let's apply this equation to our case s

Case A. The elevator moves upward so ay is positive

     F₁ = μ m (g + a_{y}) + m aₓ

Case B. The elevator is move down so that ay is negative

     F₂ = μ m (g - a_{y}) + m aₓ

    We see that F₁ > F₂

Case A requires more force than Case B

For the force to be the same the friction must be zero, which is an ideal chao since there is always some friction

1 a

2 b

3 b

4 d

5 d

6 d

explanation:

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.