A 100kg shot putter is putting a shot 4kg. Just before he releases the shot, he is in the air (feet are off the ground). At this instant, the only force he exerts against the shot is an 800N horizontal force directed forward.
a-what is his horizontal acceleration at this instant?
b-what is his vertical acceleration at this instant?
c-what is the shots horizontal acceleration at this instant?
d-what is the shots vertical acceleration at this instant?
To answer these questions, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).
a) To find the shot putter's horizontal acceleration, we need to calculate the net horizontal force acting on him. The only force in the horizontal direction is the force he exerts against the shot, which is 800N directed forward. Since there are no other horizontal forces acting on the shot putter, the net horizontal force is equal to the force he exerts.
Thus, the horizontal acceleration of the shot putter is given by a = F/m, where F is the force (800N) and m is the mass of the shot putter (100kg).
a = 800N / 100kg
a = 8 m/s^2
Therefore, the shot putter's horizontal acceleration at this instant is 8 m/s^2.
b) To find the shot putter's vertical acceleration, we need to consider the forces acting vertically. Since he is in the air, the only vertical force acting on him is the force of gravity, which is 9.8m/s^2 downward. There is no other vertical force acting on the shot putter.
Therefore, the shot putter's vertical acceleration at this instant is equal to the acceleration due to gravity (9.8m/s^2) and is directed downward.
c) To find the shot's horizontal acceleration, we need to consider the net horizontal force acting on it. The only horizontal force acting on the shot is the force exerted by the shot putter, which is 800N directed forward.
Thus, the shot's horizontal acceleration is given by:
a = F/m, where F is the force (800N) and m is the mass of the shot (4kg).
a = 800N / 4kg
a = 200 m/s^2
Therefore, the shot's horizontal acceleration at this instant is 200 m/s^2.
d) The shot does not experience any vertical forces (assuming no air resistance), so its vertical acceleration is zero. Therefore, the shot's vertical acceleration at this instant is zero.