# 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?

Oct 23, 2017

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
${F}_{\text{net}} = m \cdot a$

The net force, ${F}_{\text{net}}$, 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.

${F}_{\text{net}} = m \cdot g \cdot \sin 13 - 503 N$

${F}_{\text{net}} = 228 k g \cdot 9.8 \frac{m}{s} ^ 2 \cdot \sin 13 - 503 N = 0$

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