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Exploring Energy ConceptsDuring the packaging process, a
Chapter 2, Problem 15P(choose chapter or problem)
During the packaging process, a can of soda of mass \(0.4 \mathrm{~kg}\) moves down a surface inclined \(20^{\circ}\) relative to the horizontal, as shown in Fig. P2.15. The can is acted upon by a constant force R parallel to the incline and by the force of gravity. The magnitude of the constant force R is \(0.05 \mathrm{~N}\). Ignoring friction between the can and the inclined surface, determine the can's change in kinetic energy, in J, and whether it is increasing or decreasing. If friction between the can and the inclined surface were significant, what effect would that have on the value of the change in kinetic energy? Let \(g=9.8 \mathrm{~m} / \mathrm{s}^2\).
Questions & Answers
QUESTION:
During the packaging process, a can of soda of mass \(0.4 \mathrm{~kg}\) moves down a surface inclined \(20^{\circ}\) relative to the horizontal, as shown in Fig. P2.15. The can is acted upon by a constant force R parallel to the incline and by the force of gravity. The magnitude of the constant force R is \(0.05 \mathrm{~N}\). Ignoring friction between the can and the inclined surface, determine the can's change in kinetic energy, in J, and whether it is increasing or decreasing. If friction between the can and the inclined surface were significant, what effect would that have on the value of the change in kinetic energy? Let \(g=9.8 \mathrm{~m} / \mathrm{s}^2\).
Step 1 of 6
Here we have to calculate the change in kinetic energy when a soda of mass \(m=0.4 \mathrm{~kg}\) moves on an inclined plane \(\theta=20^{\circ}\) relative to the horizontal surface. A force \(R=0.05 N\) parallel to the inclined plane, acts on it along with the force of gravity by neglecting the friction between can and the surface. Also, we need to comment on the value of change in kinetic energy , when the friction between the can and surface becomes significant.
Given data,
Mass of can, \(m=0.4 \mathrm{~kg}\)
Parallel force, \(R=0.05 N\)
Acceleration due to gravity, \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\)
Angle, \(\theta=20^{\circ}\)
Vertical height, \(h=1.5 \mathrm{~m}\)
To find,
Change in kinetic energy, \(\Delta K E=?\)