- #1

Becca93

- 84

- 1

**Homework Statement**

A block with mass m = 7.4 kg slides down an inclined plane of slope angle 43.8 ° with a constant velocity. It is then projected up the same plane with an initial speed 3.10 m/s. How far up the incline will the block move before coming to rest?

**The attempt at a solution**

You need to find the force parallel of the block on the incline, so

Fpara = mgsinθ

Fpara = (7.4)(9.8)sin(43.8)

Fpara = 50.194 N

Acceleration down is

a = F/m

a = 50.194/7.4

a = 6.783

If the block begins moving up at Vi = 3.10 m/s, and Vf = 0, then

Vf^2 = Vi^2 + 2ad

0 = 3.1^2 + (2)(-6.783)d

d = -(3.1)^2 / -13.566

d = 0.708

This is not correct. Obviously, I'm doing this wrong, but I'm not sure of any other way to do it. I feel like having such a large acceleration down can't possibly be correct, and if that is indeed the case, how do I find acceleration of the block as it goes up the incline?