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

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A 5 kg trolley is rolling up a 25° slope with a coefficient of friction of 0.4, initially at 12 m/s. The acceleration while moving uphill is calculated as -7.85 m/s² due to the opposing forces of gravity and friction. To find the speed on the return down the slope, the distance traveled uphill is determined to be 9.17 m. The acceleration during the descent is positive, calculated using the same forces but with the direction reversed. The final speed as the trolley passes point A on its way back down can be derived from these values.
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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
 
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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

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

$v_f = \sqrt{ 0^2 + 2a \Delta x}$
 
skeeter said:
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!
 
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