Question

When a 80kg block of mass is pushed by 400 Newton at an angle of 37 degrees above the horizontal to accelerate from 20 m/s what will be the final velocity after 6 second

Answer 1

Is it pushed up or down the slope?
Is that 20 m/s up slope or down slope?
I will assume both are UP and you will have to get it straight yourself.

weight component down slope = m g sin 37
= 80 * 9.81 * sin 37 = 472 N

If you are pushing up with 400 N, you lose, at least on earth.

a = (400-472)/80 = - 0.9 m/s^2

if it started with 20 m/s UP
V = Vi -0.9 t
V = 20 - 0.9 (6)
V = 14.6 m/s

If my assumption of directions is wrong, you fix it.

Answer 2

I assume that the block of mass m=80 kg slides horizontally, and that the force F=400N pushes downwards at an angle of θ=37° below the horizontal.
Initial horizontal velocity is u=20 m/s, and we look for velocity v at time t=6 seconds.

Further assumption: friction μ=0.

Horizontal component of applied force
= Fcos(θ)
acceleration,
a=F/m=400cos(θ)/80=5cos(θ)

use kinematics equation
v=u+at
=20 m/s + 5cos(37) m/s^2 * 6
=20+23.959 m/s
=43.959 m/s

Answer 3

Ah, did not think of that :)