Question

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 26 m.

The acceleration of gravity is 9.8 m/s2 .

If the kinetic energy of the block at the bottom of the track is 3700 J, what is the work done against friction?
Answer in units of J.

Answer 1

Answer: The initial potential energy of the block is converted into kinetic energy at the bottom of the track, and some work is done against friction. We can use conservation of energy to find the work done against friction:

Initial potential energy = Final kinetic energy + Work against friction

The initial potential energy is mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the initial height of the block (which is equal to the radius of the track):

Initial potential energy = mgh = (24 kg)(9.8 m/s^2)(26 m) = 60432 J

The final kinetic energy is given as 3700 J.

Plugging these values into the equation above, we can solve for the work done against friction:

Work against friction = Initial potential energy - Final kinetic energy

= 60432 J - 3700 J

= 56732 J

Therefore, the work done against friction is 56732 J

Step-by-step explanation: