Aball whose mass is 0.4 kg hits the floor with a speed of 4 m/s and rebounds upward with a speed of 1 m/s. if the ball was in contact with the floor for 2.0 ms (2.0 × 10-3 s), what was the average magnitude of the force exerted on the ball by the floor? n calculate the magnitude of the gravitational force that the earth exerts on the ball: n

The average  magnitude of the force exerted on the ball by the floor is 600 N.

The gravitational force that Earth exerts on the ball is 3.92 N.

The given parameters;

The average  magnitude of the force exerted on the ball by the floor is calculated as follows;

The gravitational force that Earth exerts on the ball is calculated as follows;

F = mg

F = (0.4 x 9.8)

F = 3.92 N

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Explanation:

It is given that,

Mass of the ball, m = 0.4 kg

Initial speed of the ball, u = 4 m/s

Final speed of the ball, v = 1 m/s

Time of contact of the ball and the floor,

(a) Let F is the force exerted on the ball by the floor. It can be calculated using the definition of impulse. It is given by :

F = -600 N

|F| = 600 N

(b) The magnitude of the gravitational force that the Earth exerts on the ball is equal to the weight of the ball. It is given by :

W = 3.92 kg

Hence, this is the required solution.

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever