A 120 N horizontal force is applied to a 4.00 kg box on a frictionless surface. The box moves a horizontal distance of 8.00 meters. The change in the kinetic energy of the box is
A) 960 J.
B) 875 J.
C) 821 J.
D) 765 J.
E) 633 J.

Question

Quizplus · Accepted Answer

The change in kinetic energy can be found using the equation:
ΔK = 1/2 mv²f - 1/2 mv²i
where ΔK is the change in kinetic energy, m is the mass of the box, v²f is the final velocity squared and v²i is the initial velocity squared.
Since the box starts from rest, the initial velocity is 0. The final velocity can be found using the equation:
F = ma
where F is the applied force, m is the mass of the box and a is the acceleration. Since the surface is frictionless, the force applied is equal to the net force on the box. Therefore:
F = ma
a = F/m
a = 120 N / 4.00 kg
a = 30 m/s²
Using the equation of motion:
v²f = v²i + 2ad
where d is the distance traveled, we get:
v²f = 0 + 2(30 m/s²)(8.00 m)
v²f = 960
v = 31.0 m/s
Now we can find the change in kinetic energy:
ΔK = 1/2 mv²f - 1/2 mv²i
ΔK = 1/2 (4.00 kg)(31.0 m/s)² - 1/2 (4.00 kg)(0 m/s)²
ΔK = 960 J

A 120 N Horizontal Force Is Applied to a 4