An object with a mass of 10.0 kg is at rest at the top of a frictionless incline

Question

An object with a mass of 10.0 kg is at rest at the top of a frictionless inclined. plane of length 8.00 m and at an angle of 30.0degree w.r.t the horizontal. The object is released from this position and it stops at a distance d from the bottom of the inclined plane along a horizontal surface, as shown in the figure below. The coefficient of friction for the horizontal surface is 0.400. What are the speed and kinetic energy of the object at the bottom of the inclined plane? At what horizontal distance from the bottom of the plane will this object stop?

Explanation / Answer

Given that

Mass of the object m =10kg

Height of the plane h =8m

Velocity of the object at the of the plane is u=0

The speed of the object at the bottom of the inclined plane is (v)=10m/s

Now work doen by the force of friction is change in energy

Now the energy at the top is E1 =mhg =10*10*10 =1000J

And energy at the bottom of the plane is

                                          E2 =(1/2)mv2 =0.5*10*(10m/s)2 =0.5*10*100 =500

Now the change in eenrgy is W =E2-E1 =1000-500 =500J

Now the final velocity at the bottom of the ramp is given by

v2 =u2+2as

v2 =2as

The speed at the bottom of the ramp is v=Sqrt(2*gsintheta*s) =Sqrt(2*10*sin30*8) =8.944m/s

The horizontal distance moved is given by

s =v2/2a =v2/2ug

Where u is the horizontal friction

s =(8.944)2/2*0.400*10 =80/8=10m

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