FIGURE 8-8
-An object of mass 2.00 kg starts at rest from the top of a rough inclined plane of height 20.0 m as shown in Figure 8-8. If the work done by the force of friction is -150 J, what is the speed of the object as it reaches the bottom of the inclined plane? Use g = 10.0 m/s².

Question

FIGURE 8-8 FIGURE 8-8 -An object of mass 2.00 kg starts at rest from the top of a rough inclined plane of height 20.0 m as shown in Figure 8-8. If the work done by the force of friction is -150 J, what is the speed of the object as it reaches the bottom of the inclined plane? Use g = 10.0 m/s<sup>2</sup>. A) 15.8 m/s B) 150 m/s C) 10.0 m/s D) 20.0 m/s E) 200 m/s

FIGURE 8-8 -An object of mass 2.00 kg starts at rest from the top of a rough inclined plane of height 20.0 m as shown in Figure 8-8. If the work done by the force of friction is -150 J, what is the speed of the object as it reaches the bottom of the inclined plane? Use g = 10.0 m/s<sup>2</sup>. A) 15.8 m/s B) 150 m/s C) 10.0 m/s D) 20.0 m/s E) 200 m/s

-An object of mass 2.00 kg starts at rest from the top of a rough inclined plane of height 20.0 m as shown in Figure 8-8. If the work done by the force of friction is -150 J, what is the speed of the object as it reaches the bottom of the inclined plane? Use g = 10.0 m/s².

Quizplus · Accepted Answer

The potential energy at the top is converted into kinetic energy and work done against friction. Potential energy = mgh = 2 * 10 * 20 = 400 J. The work done by friction is -150 J, so the kinetic energy at the bottom is 400 - 150 = 250 J. Using KE = 0.5 * m * v^2, we solve for v: 250 = 0.5 * 2 * v^2, giving v = 15.8 m/s.

FIGURE 8-8 -An Object of Mass 2.00 Kg Starts at Rest from at Rest