Calculating Work Done on Crate by Truck

[Oliviam12]

Homework Statement

A truck carrying a 37 kg crate accelerates uniformly from rest to 77.5 km/hr in 12.2 s. Calculate the work done on the crate by the truck.

Homework Equations

W=Fcos$$\theta$$$$\Delta$$D

The Attempt at a Solution

I first changed 77.5 km/hr to 21.257 m/s. Then I multiplied 21.257 m/s by 12.2 s to get the distance which is 259.3354 m. Then I multiplied 37 kg by 9.81 m/s/s to get the force, which was 362.97 N. When I put them into the equation ( I put theta as 0) I get 94130.970 J. My teacher looked at it and said it was wrong and I am not sure what I need to do to fix it? Any ideas or something I am missing?

 

[Doc Al]

I first changed 77.5 km/hr to 21.257 m/s.

OK.

Then I multiplied 21.257 m/s by 12.2 s to get the distance which is 259.3354 m.

Incorrect: 21.257 m/s is the final speed. What's the average speed?

Then I multiplied 37 kg by 9.81 m/s/s to get the force, which was 362.97 N.

Incorrect: That's the crate's weight--you don't need that. What you need is the horizontal force that the truck exerts on the crate.

In order to find the force on the crate, use Newton's 2nd law. But you first need to figure out the acceleration using kinematics.

(You can also use the work energy theorem to solve this, but I suspect you haven't covered that yet.)