A constant 5 N horizontal force hits a stationary 2 kg ball. What is the balls speed after going 10 m?
How long was the force in contact with the ball?
Force * time = change in momentum = m v
Oh, I see, the force is forever
a = F/m = 5/2 = 2.5 m/s^2
10 = (1/2) a t^2
20 = 2.5 t^2
t = 6.93 seconds
V = a t = 2.5 *6.93 = 17.3 m/s
t = 2.83 seconds
v = 2.5 * 2.83 = 7.07 m/s
To find the ball's speed after it has traveled a distance of 10 m, we can use the concept of work and energy.
The work done on an object is defined as the force applied multiplied by the distance over which the force is applied. Mathematically, work (W) is given by:
W = F * d
In this case, the constant horizontal force applied on the ball is 5 N (newtons), and the distance traveled by the ball is 10 m (meters). So, the work done on the ball is:
W = 5 N * 10 m
W = 50 Nm (newton-meters), which is equivalent to 50 J (joules)
The work done on the ball is equal to the change in kinetic energy of the ball, as per the work-energy theorem. The change in kinetic energy (ΔKE) is given by:
ΔKE = KE_final - KE_initial
Since the ball starts from rest, the initial kinetic energy (KE_initial) is zero. Therefore, the change in kinetic energy is equal to the final kinetic energy (KE_final).
ΔKE = KE_final = 50 J
The kinetic energy of an object is defined as one-half times the mass of the object multiplied by the square of its speed. Mathematically, kinetic energy (KE) is given by:
KE = (1/2) * m * v^2
Here, the mass of the ball is 2 kg. Rearranging the formula, we can solve for the speed (v) of the ball:
v^2 = (2 * KE) / m
v^2 = (2 * 50 J) / 2 kg
v^2 = 100 J / 2 kg
v^2 = 50 m^2/s^2
Taking the square root of both sides of the equation gives us the speed (v) of the ball:
v = sqrt(50 m^2/s^2)
v ≈ 7.07 m/s
So, the ball's speed after traveling a distance of 10 m is approximately 7.07 m/s.