A 20kg box rests on a rough floor. It takes a horizontal force of 40 N to keep the box sliding at a steady speed. Suppose a force of 100 N is applied to the box. What speed would thr box have after sliding for 3.0m?
The net accelerating force is 60N.
F=ma
a= 60/20=3m/s^2
Vf^2=Vi^2+2ad Vi is zero, you know a and d, solve for Vf
To find the speed of the box after sliding for 3.0 m, we can use the concept of work and energy.
First, let's determine the work done by the 100 N force on the box. The work done is given by the equation:
Work = Force * Distance * cos(theta)
Since the force is applied horizontally, the angle theta between the force and displacement is 0 degrees. Therefore, cos(theta) = 1.
So, Work = 100 N * 3.0 m * 1 = 300 J (Joules)
Next, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. The formula is:
Work = Change in Kinetic Energy
Since the box starts from rest, it initially has zero velocity and hence zero kinetic energy.
Therefore, the initial kinetic energy (K1) is zero.
The final kinetic energy (K2) can be calculated using the formula:
K2 = (1/2) * mass * velocity^2
Given: mass (m) = 20 kg
To find the final velocity, we need the change in kinetic energy, which is equal to the work done:
K2 - K1 = Work
(1/2) * m * v^2 - 0 = 300 J
(1/2) * 20 kg * v^2 = 300 J
10 kg * v^2 = 300 J
v^2 = 30 J / 10 kg
v^2 = 30 m^2/s^2
v = √(30) = 5.48 m/s (approximately)
Therefore, the speed of the box after sliding for 3.0 m would be approximately 5.48 m/s.