Starting from rest at a height equal to the radius of the circular track, a block of mass 24 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient µ). the radius of the track is 15 m. the acceleration of gravity is 9.8 m/s 2 .starting from rest at a height equal to the radius of the circular track, a block of mass 24 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient µ). the radius of the track is 15 m. the acceleration of gravity is 9.8 m/s 2 . if the kinetic energy of the block at the

bottom of the track is 3900 j, what is the

work done against friction?

answer in units of j.

The work done against friction is 372 joules

Explanation:

It is given that,

Mass of block, m = 24 kg

Radius of the track, r = 15 m

Acceleration due to gravity,

If the kinetic energy of the block at the  bottom of the track is, 3900 J

Let P is the work done against friction. It is given by :

Here, h = r

P = 3528 J

Since it ends up with 3900 J, the work done is given by or the lost in energy will be :

W = 3900 - 3528

W = 372 joules

So, the work done against friction is 372 joules. Hence, this is the required solution.

t2 equal the displacement of the particle over the same time period if t2> t/2? hint: consider a sine wave starting at t1=0, then consider a closing wave starting at t1=0, for your explanatio

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