Mass on plane, applied force, and kinetic friction

[IAmPat]

Homework Statement

A block of mass 15.0 kg sits on a plane that is inclined at a 37.0° angle to the horizontal. A 227 N force, pointing up the plane, is applied to the block. The coefficient of kinetic friction is 0.500. What is the speed of the block after 2.00 s?

Homework Equations

m = 15.0kg
Fn = 147N
$$\mu$$k = .500

The Attempt at a Solution

Fk = 73.5N (natural force * kinetic friction coefficient)
$$\sum$$Fx = 227 - 73.5 = 153.5
153.5 = (15)$$a_{x}$$
$$a_{x}$$ = 10.23I got to here and then stopped because I didn't see how I could work out a solution using this method. Is there another equation I can use, or a different approach I should take?

 

[anathema]

Hi there! It looks like you're on the right track with your solution so far. To find the speed of the block after 2.00 s, you can use the equation v = u + at, where v is the final velocity, u is the initial velocity (which is 0 in this case), a is the acceleration, and t is the time.

Since you have already calculated the acceleration (a) to be 10.23 m/s^2, you can plug that into the equation along with the given time of 2.00 s to find the final velocity (v).

v = 0 + (10.23 m/s^2)(2.00 s)
v = 20.46 m/s

Therefore, the speed of the block after 2.00 s is 20.46 m/s. I hope this helps! Let me know if you have any further questions.