If a pitcher throws a fast ball with a speed of 35 m/s. The ball has a mass of 220 grams. a) how much work was done in throwing the ball? b) If the catcher catches the ball by moving his mitt 15 cm. backward what force did he exert?
a.) Throwing the ball in a positive direction
v= 35 m/s
m= .220 kg
wk= KE= 1/2 mv^2
wk= (1/2)(.220)(35)^2= 1.35x10^2
b.)When the pitcher moves mitt backwards, that would be a negative distance? d= -0.15 m.
F.d= change in KE
-0.15F= 0-((1/2)(0.220)(35)^2))
F= 8.98x10^2 N
All correct!
To solve both parts of the problem, we need to use the formulas for work and force.
a) The work done when throwing the ball can be calculated using the formula:
Work (W) = Kinetic energy (KE) = 1/2 * mass (m) * velocity squared (v^2)
In this case, the mass (m) of the ball is 220 grams, which can be converted to kilograms by dividing by 1000:
m = 220 g / 1000 = 0.220 kg
The velocity (v) of the ball is given as 35 m/s. Plugging these values into the formula:
W = (1/2) * 0.220 * (35)^2
W = 0.5 * 0.220 * 1225
W = 134.75 Joules
Therefore, the work done in throwing the ball is 134.75 Joules.
b) To find the force exerted by the catcher when moving his mitt backward, we can use the formula:
Work (W) = Force (F) * distance (d)
Given that the distance (d) moved by the mitt is 15 cm, we need to convert it to meters by dividing by 100:
d = 15 cm / 100 = 0.15 m
The work done is equal to the change in kinetic energy (KE) of the ball. In this case, the ball comes to rest after being caught, so its final kinetic energy is zero:
W = change in KE = 0 - (1/2 * mass * velocity^2)
Plugging in the values:
0.15F = 0 - (1/2 * 0.220 * 35^2)
Solving for force (F):
F = -(1/0.15) * (0.5 * 0.220 * 1225)
F = -0.15 * 133.75
F = -19.93 N (approximately -20 N)
Therefore, the force exerted by the catcher is approximately -20 Newtons, with the negative sign indicating that the direction of the force is opposite to the movement of the ball.