A 2.0-kg block slides down a fixed,rough curved track.The block has a speed of 5.0 m/s after its height above a horizontal surface has decreased by 1.8 m.Assume the block starts from rest.What is the change in mechanical energy of the block caused by the force of friction during this descent?
A)-14 J
B)-12 J
C)-10 J
D)-16 J
E)-25 J

Question

Quizplus · Accepted Answer

The change in mechanical energy is equal to the work done by all forces acting on the object. In this case, the only force doing work is friction, which acts against the direction of motion, slowing the block down. The work done by friction is equal to the force of friction times the distance traveled, which can be calculated using the work-energy principle:

W_friction = ΔK = K_f - K_i

where W_friction is the work done by friction, ΔK is the change in kinetic energy, K_f is the final kinetic energy (when the block reaches the bottom of the track), and K_i is the initial kinetic energy (when the block starts from rest).

We can use the kinematic equations to find the final velocity of the block at the bottom of the track:

v_f^2 = v_i^2 + 2gh
v_f^2 = 0 + 2(9.8 m/s^2)(1.8 m)
v_f = 5.4 m/s

Then we can calculate the initial kinetic energy:

K_i = (1/2)mv_i^2 = 0

And the final kinetic energy:

K_f = (1/2)mv_f^2 = (1/2)(2.0 kg)(5.4 m/s)^2 = 29.16 J

Finally, we can calculate the work done by friction:

W_friction = K_f - K_i = 29.16 J - 0 J = 29.16 J

Therefore, the change in mechanical energy caused by the force of friction during the descent is -29.16 J (negative because friction does negative work). The closest answer choice to this value is choice C, which is 10 J off. None of the other choices are within a reasonable range.

A 20-Kg Block Slides Down a Fixed,rough Curved Track