a metal crate with a mass of 500 kg is pushed across a metal floor for 5 minutes.50N is applied horizontally.box does not move, extra 50 N is applied.after 2 minutes then box moves 4meters,how much work is done
no work is done until the box moves.
W = F * d
W = 100 N * 4 meters = 400 Joules
Most of this question is added nonsense intended to confuse you.
We push a large box for three minutes. During that time, we exert a constant force of 200N on the box, but it does not move. How much work has been accomplished?
NONE !!
Force times distance moved in the direction of the force !!!!!
To calculate the work done, we need to first calculate the force required to overcome the friction between the crate and the floor. Once we have the force, we can use it together with the distance moved to calculate the work done.
Given:
Mass of the crate (m) = 500 kg
Force applied initially (F1) = 50 N
Additional force applied (F2) = 50 N
Distance moved (d) = 4 meters
Time taken to move (t) = 2 minutes = 120 seconds
Step 1: Calculating the force of friction
Initially, the crate does not move, so the applied force is equal to the force of friction. Thus:
F1 = Force of friction
Step 2: Calculating the acceleration
Using Newton's second law of motion, we can calculate the acceleration (a) of the crate:
F1 = m * a
Step 3: Calculating the force required after 2 minutes
After applying an additional force (F2 = 50 N), the crate moves with a constant velocity. This means that the net force on the crate is zero, as it overcomes the force of friction. Thus:
F2 = Force of friction
Step 4: Calculating the acceleration after 2 minutes
Using the formula again:
F2 = m * a
Step 5: Calculating the work done
The work done (W) is given by the formula:
W = Force * Distance
Now let's calculate the values:
Step 1:
F1 = 50 N
Step 2:
a = F1 / m
= 50 N / 500 kg
= 0.1 m/s^2
Step 3:
F2 = 50 N
Step 4:
a = F2 / m
= 50 N / 500 kg
= 0.1 m/s^2
Step 5:
W = F2 * d
= 50 N * 4 meters
= 200 Joules
Therefore, the work done to move the crate a distance of 4 meters is 200 Joules.