A man accelerates a 25 kg box by pushing it with a force of 60 N over a distance of 10.0 m on a surface that provides a frictional force of 10 N
Determine the velocity of the box after the man stops pushing in two ways
a) by calculating net force,the acceleration and then the average speed
b by cacualating net work done on the box and then considering the changes in energy of the box
Net force = F = 60 - 10 = 50 Newtons
F = m a
50 = 25 a
a = 2 m/s^2
v = a t = 2 t (assuming v =zero at t = 0)
x = (1/2) a t^2 = t^2
10 = (1/2)(2) t^2
so t = sqrt (10)
speed at 10 meters = 2 sqrt 10 = ANSWER final speed
so average speed = sqrt (10)
work done= F d = 50*10 = 500 Joules
(1/2) m v^2 = 500
(1/2)(25) v^2 = 500
v^2 = 1000/25 = 40
v = sqrt( 4*10) = 2 sqrt (10) [ whew, same answer ]
To determine the velocity of the box after the man stops pushing, we can use two different approaches:
a) Calculating net force, acceleration, and average speed:
1. Net force: To calculate the net force acting on the box, we need to subtract the force of friction from the applied force. In this case, the applied force is 60 N, and the frictional force is 10 N. Therefore, the net force is 60 N - 10 N = 50 N.
2. Acceleration: We can use Newton's second law, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. The mass of the box is given as 25 kg. Hence, the acceleration can be calculated as acceleration = net force / mass = 50 N / 25 kg = 2 m/s^2.
3. Average speed: The average speed is the total distance traveled divided by the total time taken. In this case, the box travels a distance of 10.0 m. To find the time taken, we need to calculate the time using the kinematic equation: distance = (initial velocity * time) + (0.5 * acceleration * time^2). Since the initial velocity is 0 (as the man stops pushing), we can simplify this equation to: time = sqrt((2 * distance) / acceleration) = sqrt((2 * 10.0 m) / 2 m/s^2) = sqrt(20) = 4.47 s. Now we can calculate the average speed by dividing the distance by the time: average speed = distance / time = 10.0 m / 4.47 s = 2.24 m/s.
b) Calculating net work done on the box and considering changes in energy:
1. Net work: The net work done on an object is equal to the change in its kinetic energy. In this case, the initial velocity is 0 (as the man stops pushing), so the initial kinetic energy is 0. The final kinetic energy can be calculated using the formula: final kinetic energy = (1/2) * mass * velocity^2. Since we're trying to find the final velocity, we rearrange this formula to: velocity = sqrt((2 * final kinetic energy) / mass). Initially, the box is at rest, so the net work done on the box is equal to the work done by the applied force, which is work = force * distance = 60 N * 10.0 m = 600 J.
2. Changes in energy: The change in kinetic energy is equal to the work done on the box. Therefore, the final kinetic energy is equal to the work done, since the initial kinetic energy is 0. Substituting the values, we get: final kinetic energy = 600 J. Now we can calculate the final velocity using the formula from the previous step: velocity = sqrt((2 * 600 J) / 25 kg) = sqrt(48) = 6.93 m/s.
So, the velocity of the box after the man stops pushing is 2.24 m/s (calculated using net force and average speed) and 6.93 m/s (calculated using net work and changes in energy).