A 100N force is applied to a 10kg object at an angle of 30 degrees for 2 seconds. If the object is initially at rest, what is its final velocity? (ignore friction: sin 30 degrees= 0.5; cos 30 degrees= 0.87)

A. 8.7 m/s

B. 1 m/s

C. 17.4 m/s

D. 34.8 m/s

I solved this equation using the W=K. I figured that there was no friction so there was no heat and no change in internal energy. The object is also on a flat surface with no incline so there is no potential energy to worry about and all the energy is transferred into Kinetic Energy.

So W=K, W=Fdcosθ, K= ½mv2

Thus, Fdcosθ= ½mv2

In addition, d=tv

F(tv)cosθ= ½mv2 simplified is F(t)cosθ= ½mv

Now we can plug and chug F=100N, t=2s, θ=30 degrees, m=10kg, and we are solving for v

(100N)(2s)cos30= ½(10kg)v

**v=34.8 m/s**which is answer

**d**

**Their explanation is The horizontal component of the force is 87N. To find the acceleration we use F=ma. a=8.7 m/s2. We use v=vo+at and arrive at 17.4 m/s.**

*Except, according to exam krackers the answer is c.*I understand how they solved it. But why is the way I solved it wrong. What did I do wrong? Can someone explain this to me.

Thank you,

Veronica