A mass of 11 kg is moving at a speed of 20 m/s at the bottom of incline which is

Question

A mass of 11 kg is moving at a speed of 20 m/s at the bottom of incline which is inclined at an angle of 61.2 degrees. The incline has friction and the coefficient of kinetic friction is 0.72. The mass travels up the incline and stops due to both gravity and friction and then slides back down the incline. When the object slides back down the incline, how fast is it moving at the bottom in m/s?

Please show work. The correct answer is 13.16 but I got 16.58 (which is wrong)

A mass of 11 kg is moving at a speed of 20 m/s at the bottom of incline which is inclined at an angle of 61.2 degrees. The incline has friction and the coefficient of kinetic friction is 0.72. The mass travels up the incline and stops due to both gravity and friction and then slides back down the incline. When the object slides back down the incline, how fast is it moving at the bottom in m/s?

Please show work. The correct answer is 13.16 but I got 16.58 (which is wrong)

Explanation / Answer

Considering the free body diagram we get acceleration of a body moving up the ramp is given by

a =(gsintheta+Fk)m

But the frictional force is given by Fk =uN =u(mgcostheta)

Now the acceleration is a =gsintheta+ugcostheta

=9.81(sin61.2+0.72*cos(61.2))

=9.81(0.876+0.346) =11.996m/s2

The distance moved by the body up the ramp is given by

v2-u2 =2as Final velocity becomes v =0

-u2 =2*-11.996*s

(20)2 =2*11.996*s ====>s =16.672m

After this the body moves down the ramp then the acceleration is given by

a =g(sintheta -ucosthea)

=9.81(sin61.2+0.72*cos(61.2))

=9.81(0.876-0.346) =5.12m/s2

When the object slides back down the incline, how fast is it moving at the bottom in m/s is given by

v2-u2 =2as

v2 =u2+2as ===> v=Sqrt(2g*s) =Sqrt(2*5.12*16.672) =13.06m/s almost nearly equal to your answer

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