Question

To unload the truck, the man pushes the crate down the ramp, which is now angled at 13°, using a force of 503N. What is the acceleration of the crate down the ramp?

Answer 1

It depends on information not available.

Explanation:

What is the mass of the crate?
In what direction is the 503 N force exerted?
Is friction to be considered?

As an example, I will choose information to be added and then give the acceleration:
The mass of the crate is 228 kg. The direction of the applied 503 N force is up the ramp. Ignore friction, the crate is on massless wheels.

The acceleration will be given by Newton's 2nd Law
`F_"net" = m*a`

The net force, `F_"net"`, is the vector sum of the component of the weight of the crate, m*g, that points down the ramp and the 503 N force that points up the ramp.

`F_"net" = m*g*sin13 - 503 N`

`F_"net" = 228 kg*9.8 m/s^2*sin13 - 503 N = 0`

So the net force on the crate is zero. Therefore the acceleration is zero. The man forces the crate to go down the ramp at constant speed because there is a right turn he must make at the bottom of the ramp.

I hope this helps,
Steve