A truck carrying a 57 kg crate accelerates uniformly from rest to 86.5 km/hr in 18.4 s. Calculate the work done on the crate by the truck (in J).
I got 16454 J right. Now why is it positive?
K-K = negative if Velocity from rest is 0.
no, if you push something to get it going you do positive work on it. Your force is in the direction of motion.
I you were pushing a spring, it would do negative work on you.
-(1/2)k x^2 :)
but you would do positive work on it.
To calculate the work done on the crate by the truck, we need to use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy.
First, let's convert the final velocity from km/hr to m/s. We know that 1 km/hr is equal to 0.2778 m/s. So, 86.5 km/hr is equal to 86.5 * 0.2778 m/s = 24.05 m/s.
Next, we can calculate the initial velocity of the crate. Since the truck starts from rest, the initial velocity is 0 m/s.
Now, we can calculate the change in kinetic energy of the crate using the formula:
ΔK.E. = (1/2) * mass * (final velocity^2 - initial velocity^2)
Plugging in the values:
ΔK.E. = (1/2) * 57 kg * (24.05 m/s)^2 - (0 m/s)^2
Simplifying:
ΔK.E. = (1/2) * 57 kg * (578.4025 m^2/s^2)
ΔK.E. = 16,454.5625 J
Now, why is the work done on the crate considered positive?
In this case, the truck is accelerating the crate in the same direction as its motion. When work is done on an object, it means energy is being transferred to that object. Since the truck is supplying the energy to accelerate the crate, the work done on the crate is positive.
If the truck were decelerating or moving in the opposite direction of the crate, the work done on the crate would be negative because the truck would be taking energy away from the crate.