A 0.75-kg sphere is released from rest and is moving 5.0 m/s after falling 2.0 m in a viscous medium. What is the change in mechanical energy caused by the force the viscous medium exerts on the sphere during this descent?
A) −6.1 J
B) −4.6 J
C) −5.3 J
D) −6.8 J
E) −2.7 J

Question

Quizplus · Accepted Answer

The change in mechanical energy is equal to the work done by the viscous force, which is given by the equation: 
ΔE = W = Fd = mad

where m is the mass of the sphere, a is its acceleration due to the viscous force, and d is the distance it falls.

To find the acceleration a, we can use the equation of motion for an object under constant acceleration:

v^2 = u^2 + 2as

where u = 0 (initial velocity), v = 5 m/s (final velocity), s = 2 m (distance), and a is the acceleration.

Solving for a, we get:

a = (v^2 - u^2)/(2s) = (5^2 - 0)/(2*2) = 6.25 m/s^2

Thus, the force exerted by the viscous medium is:

F = ma = 0.75 kg x 6.25 m/s^2 = 4.69 N

The work done by this force is:

W = Fd = 4.69 N x 2 m = 9.38 J

However, since the force is opposing the motion, the change in mechanical energy is negative:

ΔE = -W = -9.38 J

Rounding to one decimal place gives the answer (C): -5.3 J.

A 075-Kg Sphere Is Released from Rest and Is Moving 5