Answer:
A.) 12.5 J
B.) 12.5 J
C.) 7.1 m/s
Explanation:
Given that a 0.5 kg object, initially at rest, is pulled to the right along a frictionless horizontal surface by a constant horizontal force of 25 N for a distance of 0.5m.
a. What is the work done by the force?
Work done = force × distance
Work done = 25 × 0.5
Work done = 12.5 J
b. What is the change in the kinetic energy of the block?
Work done = energy
Change in Kinetic energy = work done
Change in kinetic energy = 12.5 J
c. What is the speed of the block after the force is removed?
Kinetic energy = 1/2mV^2
12.5 = 1/2 × 0.5 × V^2
25 = 0.5V^2
V^2 = 25/0.5
V^2 = 50
V = 7.1 m/s
