# Really simple problem - can't get right answer!

1. Sep 25, 2004

### justagirl

Suppose a 50kg block slides along a
horizontal surface where the coefficient of kinetic friction
between the block and the surface is uk = 0.60. A force F =
400 N is now applied, where the
angle of the force above horizontal is 20°.

What is the magnitude of the acceleration of the block?

0.54 m/s2
2.31 m/s2
3.27 m/s2
6.78 m/s2
8.11 m/s2

Here's what i did:

F(x-net) = F(applied) - F(friction) = ma
a = F(applied) - F(friction) / m
= (400)(cos 20) - (50 * 9.8 * 0.6) / 50
= 375 N - 294 N / 50 kg
= 1.67 m/s2

I checked my numbers quite a few times and keep getting
1.67, which is not one of the answer choices. The only thing
I saw was that 3.72 m/s2 is exactly double what I got, but I
can't find an explanation for that.

Thanks!

2. Sep 25, 2004

### marlon

The answer is 3.27.
You gave the right anwser for the acceleration in the x-direction. Now when you look vertically in the y direction you still have the y-component of the 400 N-force that you forgot (gravity and normal force annihilate each other)

This y-component = 400*sin(20)

Then calculate a_y and you already have a_x =1.67 which is correct.

The magnitude is sqrt((a_x)² + (a_y)²)

marlon