Solving Friction Problem: Mass 5kg Trolley, 25° Slope, Coefficient 0.4

[Shah 72]

A trolley of mass 5 kg is rolling up a rough slope, which is at an angle of 25 degree to the horizontal. The coefficient of friction between the trolley and the slope is 0.4. It passes a point A with speed 12m/s. Find its speed when it passes A on its way back down the slope.
So I did F=m×a
-0.4×50cos25- 50sin25= 5a
a=-7.85m/s^2.
I don't know how to calculate after this. Pls help

 

[skeeter]

let uphill be positive …

initial velocity > 0, acceleration < 0

$a = -g(\sin{\theta} + \mu\cos{\theta})$

$\Delta x = \dfrac{0^2-12^2}{2a} = 9.17 \, m$let downhill be positive …

initial velocity = 0, final velocity > 0, acceleration > 0, delta x > 0

$a = g(\sin{\theta} - \mu \cos{\theta})$

$v_f = \sqrt{ 0^2 + 2a \Delta x}$

 

[Shah 72]

let uphill be positive …

initial velocity > 0, acceleration < 0

$a = -g(\sin{\theta} + \mu\cos{\theta})$

$\Delta x = \dfrac{0^2-12^2}{2a} = 9.17 \, m$let downhill be positive …

initial velocity = 0, final velocity > 0, acceleration > 0, delta x > 0

$a = g(\sin{\theta} - \mu \cos{\theta})$

$v_f = \sqrt{ 0^2 + 2a \Delta x}$

Thank you so much!