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[tex]\theta[/tex][tex]\Delta[/tex]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.)

 

[ogb p]

I would like to commend you for your attempt at solving this problem. Your approach is correct, but there may have been some calculation errors. Let's break down the steps to find the work done on the crate by the truck.

Step 1: Convert the initial velocity from km/hr to m/s
You correctly converted 77.5 km/hr to 21.257 m/s.

Step 2: Calculate the distance traveled by the crate
You multiplied the initial velocity by the time, which is correct. However, the time given in the problem is 12.2 s, not 12.2 m. The correct distance traveled by the crate would be 259.3354 m.

Step 3: Calculate the force applied by the truck on the crate
You correctly multiplied the mass of the crate by the acceleration due to gravity. However, to find the force applied by the truck, you need to use the equation F=ma, where "a" is the acceleration of the truck. The truck is accelerating from rest (0 m/s) to 21.257 m/s in 12.2 s, so the acceleration would be (21.257 m/s)/(12.2 s) = 1.743 m/s^2. Using this value for acceleration, the force applied by the truck on the crate would be 37 kg * 1.743 m/s^2 = 64.491 N.

Step 4: Calculate the work done on the crate
Now that we have the correct values for distance and force, we can plug them into the equation for work: W = FcosθΔD. In this case, we can assume that the force and displacement are in the same direction (θ = 0), so the equation becomes W = FΔD. Plugging in the values we calculated, we get W = (64.491 N)(259.3354 m) = 16724.9 J.

Therefore, the work done on the crate by the truck is 16724.9 J. I hope this helps clarify any confusion and guides you in the right direction for future problem-solving. Keep up the good work!