A small block of mass .15 kg is placed at top of ramp that is 2 m high It slides

Question

A small block of mass .15 kg is placed at top of ramp that is 2 m high It slides w/o friction down track and around a loop of radius .6m and leaves the loop to travel along track that is .5 meters above ground It leaves track at that height. Calculate speed of block at top of loop. Calculate speed of block when it leaves track. Determine maximum height above ground attained by mass after it leaves track. Another track is used that has friction. This track is 2 m high and the block leaves this track with velocity of 4 m/s How much work was done by frictional force

Explanation / Answer

Here,

mass of block , m = 0.15 Kg

height of ramp , h = 2 m

radius of loop , r =0.6 m
height of track , h1 = 0.5 m

let the speed of loop is vt

Using conseravtion of energy

0.5 * m * v^2 = m * g * (h - r)

0.5 * v^2 = 9.8 * (2 - 0.6)

v = 5.24 m/s

the speed of the block it leaves track is 5.24 m/s

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at the track ,

Using conseravtion of energy

0.5 * m * v^2 = m * g * (h - h1)

0.5 * v^2 = 9.8 * (2 - 0.5)

v = 5.422 m/s

the speed of the block it leaves track is 5.422 m/s

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let the maximum height of track is hmax

hmax = 2 m

maximum height above ground attained by mass after it leaves track is 2 m

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work done by friction = change in mechanical energy

work done by friction = - 0.15 * 9.8 * 2 + 0.5 * 0.15 * 4^2

work done by friction = - 1.74 J

the work done by friction is -1.74 J

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